Work Energy Power And Collisions Multiple Choice Question And Answers

Work Energy Power And Collisions

Question 1. A particle moves from a point \((-2 \hat{i}+5 \hat{j}) \text { to }(4 \hat{j}+3 \hat{k})\) when a force of \((4 \hat{i}+3 \hat{j})\)N is applied. How much work has been done by this force?

1. 8J
2. 11J
3. 5J
4. 2J

Answer: 3. 5J

The position vector of the initial position of the particle is,

⇒ \(\vec{r}_1=-2 \hat{i}+5 \hat{j}\)

and that of the final position is,

⇒ \(\overrightarrow{r_2}=4 \hat{j}+3 \hat{k}\).

∴ the displacement vector is \(\vec{s}=\vec{r}_2-\vec{r}_1=(2 \hat{i}-\hat{j}+3 \hat{k}) \mathrm{m}\).

Force \(\vec{F}=(4 \hat{i}+3 \hat{j}) \mathrm{N}\)

∴ work done is \(W=\vec{F} \cdot \vec{s}=(4 \hat{i}+3 \hat{j}) \mathrm{N} \cdot(2 \hat{i}-\hat{j}+3 \hat{k}) \mathrm{m}\)

= (8 – 3) J

= 5 J

work energy and power neet questions

Question 2. A body of mass 1 kg is thrown upwards with a velocity of 20ms^-1. It momentarily comes to rest after attaining a height of 18m. How much energy is lost due to air friction? (Take g = 10ms^-2).

1. 30J
2. 40J
3. 10J
4. 20J

Answer: 4. 20J

Work done by gravity = mgh = (1 kg)(-10 m s^-2)(18 m)

= -180 J.

Initial \(\mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2}(1 \mathrm{~kg})\left(20 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\)

= 200 J.

By the work-energy theorem,

⇒ \(W_{\text {grav }}+W_{\text {air }}=\mathrm{KE}_{\mathrm{f}}-\mathrm{KE}_{\mathrm{i}}\)

=> -180 J + W_air = 0J -200J.

∴ work done by air friction = 180 J- 200 J

= 20 J.

Hence, the energy lost due to air friction is 20 J.

Question 3. A child is swaying on a swing. Its minimum and maximum heights from the ground are 0.75 are 2.0m respectively. Its maximum speed will be

1. 10 m s^-1
2. 5 m s ^-1
3. 8 m s^-1
4. 15 m s^-1

Answer: 2. 5 m s ^-1

At the maximum height (h₂= 2.0m), the KE is zero, and at the minimum height (h₁ = 0.75 m), the KE is the maximum

work energy and power neet questions

⇒ \(\left(=\frac{1}{2} m v_{\max }^2\right)\)

Hence, by the principle of conservation of energy

⇒ \(\frac{1}{2} m v_{\max }^2=m g\left(h_2-h_1\right)\)

⇒ \(v_{\max }=\sqrt{2 g\left(h_2-h_1\right)}=\sqrt{2\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(1.25 \mathrm{~m})}\)

= 5ms^-1

Question 4. A particle having a kinetic energy K is projected making an angle of 60 with the horizontal. The kinetic energy at the highest point will be

1. K
2. \(\frac{K}{2}\)
3. \(\frac{K}{4}\)
4. \(\frac{K}{8}\)

Answer: 3. \(\frac{K}{4}\)

The horizontal component of the velocity of projection remains unchanged.

At the highest point, the velocity is along the horizontal and equals u cos 60°= \(\frac{u}{2}\)

Initial kinetic energy

⇒ \(K=\frac{1}{2} m u^2\)

At the highest point,

⇒ \(K^{\prime}=\frac{1}{2} m\left(\frac{u}{2}\right)^2=\frac{1}{2} m u^2 \times \frac{1}{4}=\frac{K}{4}\)

Question 5. A body moves through a distance of 10 m long in a straight line under the action of a 5-N force. If the work done is 25J, the angle between the force direction of the body is

1. 30°
2. 45°
3. 60°
4. 75°

Answer: 3. 60°

Work done = W = Fs cos θ.

∴ 25 J= (5N)(10 m)cos θ

⇒ \(\cos \theta=\frac{25 \mathrm{~J}}{50 \mathrm{~J}}=\frac{1}{2}\)

θ = 60°.

Question 6. A force on a 3-g particle in such a way that the position of the particle as a function of time is given by \(x = 3t- At² + t³\), where x is in meters and t is in seconds. The work done during the first four seconds is

1. 490 3J
2. 450 mJ
3. 528 mJ
4. 530 mJ

Answer: 3. 528 mJ

Given that position x = 3t- At² + t³.

So, velocity \(0=\frac{d x}{d t}=3-8 t+3 t^2\)

Initial velocity \(u=\left.\frac{d x}{d t}\right|_{t=0}=3 \mathrm{~m} \mathrm{~s}^{-1}\)

Final velocity

⇒ \(v=\left.\frac{d x}{d t}\right|_{t=4 \mathrm{~s}}=(3-32+48) \mathrm{m} \mathrm{s}^{-1}=19 \mathrm{~m} \mathrm{~s}^{-1}\)

∴ the change in kinetic energy is

⇒ \(\Delta \mathrm{KE}=\frac{1}{2} m\left(v^2-u^2\right)=\frac{1}{2}(3 \mathrm{~g})\left[\left(19 \mathrm{~m} \mathrm{~s}^{-1}\right)^2-\left(3 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\right]\)

= 528mJ

Question 7. Two bodies with their kinetic energies in the ratio 4:1 are moving with equal linear momenta. The ratio of their masses is

1. 4: 1
2. 1: 2
3. 2: 1
4. 1: 4

Answer: 4. 1: 4

Let m₁ and m₂ be the masses of the bodies and p be their equal momentum.

Now, \(\mathrm{KE}=\frac{1}{2} m v^2=\frac{m^2 v^2}{2 m}=\frac{p^2}{2 m}\)

Given that,

⇒ \(\frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{4}{1}=\frac{\frac{p^2}{2 m_1}}{\frac{p^2}{2 m_2}}=\frac{m_2}{m_1}\)

⇒\(\frac{m_1}{m_2}=\frac{1}{4} \Rightarrow m_1: m_2\)

= 1: 4

work energy and power neet questions

Question 8. Two bodies of masses m and 4m are moving with equal kinetic energies. The ratio of their linear momenta is

1. 1: 2
2. 1: 4
3. 4: 1
4. 1: 1

Answer: 1. 1: 2

Given that m₁ = m, m₂ = 4m and K₁ = K₂

⇒ \(\frac{p_1^2}{2 m_1}=\frac{p_2^2}{2 m_2} \Rightarrow \frac{p_1}{p_2}=\sqrt{\frac{m_1}{m_2}}=\sqrt{\frac{m}{4 m}}=\frac{1}{2}\)

= 1: 2

Question 9. A body constrained to move along the y-axis is subjected to a force given by \(\vec{F}=(-2 \hat{i}+15 \hat{j}+6 \hat{k})\)N. The work done by this force in moving the body through a distance of \(10 \hat{j}\) m along the y-axis is

1. 160J
2. 20J
3. 150J
4. 190J

Answer: 3. 150J

Work done \(W=\vec{F} \cdot \vec{s}=(-2 \hat{i}+15 \hat{j}+6 \hat{k}) \mathrm{N} \cdot(10 \hat{j}) \mathrm{m}\)

= 150J

Question 10. A position-dependent force F=(7- 2x + 3X²)N acts on a small body of mass 2kg and displaces it from x= 0 to x= 5m. The work done is

1. 35 J
2. 70 J
3. 135 J
4. 270 J

Answer: 3. 135 J

The work done by the given variable force is

⇒ \(W=\int d W=\int F d x=\int_0^{5 \mathrm{~m}}\left[\left(7-2 x+3 x^2\right) \mathrm{N}\right] d x\)

⇒ \(\left[\left(7 x-x^2+x^3\right) \mathrm{N}\right]_0^{5 \mathrm{~m}}=(35-25+125) \mathrm{J}\)

= 135J

Question 11. A bullet of mass 10 g leaves a rifle with an initial velocity of 1000m s^-1 and strikes the ground at the same horizontal level with a velocity of 500m s^-1. The work done in overcoming the air resistance will be

1. 375 J
2. 500 J
3. -3750 J
4. 5000 J

Answer: 3. -3750 J

By the work-energy theorem, the work done by the bullet in overcoming the air resistance is

W = change in \(\mathrm{KE}=\frac{1}{2} m v^2-\frac{1}{2} m u^2\)

⇒ \(\frac{1}{2}(10 \mathrm{~g})\left[\left(500 \mathrm{~m} \mathrm{~s}^{-1}\right)^2-\left(10^3 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\right]\)

Question 12. The kinetic energy acquired by a mass m in traveling through a distance d, starting from rest, under the action of a constant force is directly proportional to

1. m
2. m°
3. √m
4. \(\frac{1}{\sqrt{m}}\)

Answer: 2. m°

v² = u² + 2as

= 2as = 2ad [here, u = θ and s = d].

∴ \(\mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2} m(2 a s)\)

= mas

= F.s

Hence, the KE acquired is independent of m and is thus proportional to m°

Question 13. Water falls from a height of 60 m at the rate of 15 kg s^-1 to operate a turbine. The losses due to frictional forces are 10% of the energy. How much power is generated by the turbine? (Takeg = 10m s^-2.)

1. 12.3 kW
2. 7.0 kW
3. 8.1 kW
4. 10.2 kW

Answer: 3. 8.1 kW

The mass of water falling per second is 15 kg, the height is 60 m, and g = 10 ms^-2.

Now, the loss of energy is 10%.

∴ available energy = 90%.

∴ power generated = \(\left(15 \mathrm{~kg} \mathrm{~s}^{-1}\right)\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(60 \mathrm{~m})\left(\frac{90}{100}\right)\)

= 8100W

= 8.1 W.

work energy and power neet questions

Question 14. An engine pumps water continuously through a hose. Water leaves the hose with a velocity of v, and m is the mass per unit length of the water jet. What is the rate at which the kinetic energy is imparted to the water?

1. mv³
2. \(\frac{1}{2} m^2 v^2\)
3. \(\frac{1}{2} m v^2\)
4. \(\frac{1}{2} m v^3\)

Answer: 4. \(\frac{1}{2} m v^3\)

Given that velocity of water = v and mass flowing per unit length = m. In one second, the length of the water jet is v, so the mass contained is mv.

∴ the KE imparted to water in one second is

⇒ \(\frac{1}{2}(m v) v^2=\frac{1}{2} m v^3\)

Question 15. A block of mass M is attached to the lower end of a vertical spring, which is hung from the ceiling and has a force constant of k. The block is released from rest with the initial stretch. The maximum extension produced in the length of the spring will be

1. \(\frac{M g}{2 k}\)
2. \(\frac{2 M g}{k}\)
3. \(\frac{4 M g}{k}\)
4. \(\frac{M g}{k}\)

Answer: 2. \(\frac{2 M g}{k}\)

Let x be the extension in the spring.

So, PE lost by the block = PE gained by the spring

⇒ \(M g x=\frac{1}{2} k x^2 \Rightarrow x=\frac{2 M g}{k}\)

Question 16. A block2kof mass m, starting from rest, undergoes a uniform acceleration. If the speed acquired in a time t be v, the power delivered to the block is

1. \(\frac{m v^2}{t}\)
2. \(\frac{m v^2}{2 t^2}\)
3. \(\frac{m v^2}{t^2}\)
4. \(\frac{m v^2}{2 t}\)

Answer: 4. \(\frac{m v^2}{2 t}\)

Power delivered = \(\frac{\text { work done }}{\text { time }}=\frac{\text { change in the } \mathrm{KE}}{t}\)

⇒ \(\frac{\frac{1}{2} m v^2}{t}=\frac{m v^2}{2 t}\)

Question 17. A block of mass 10 kg, moving along the x-axis with a constant speed of 10 m s^-1, is subjected to a retarding force F = -0.1x J m^-1 during its travel from x = 20 m to x = 30 m. Its final kinetic energy will be

1. 275 J
2. 450 J
3. 475J
4. 250 J

Answer: 3. 475J

By the work-energy theorem, work done by the retarding force = change in the KE.

∴ \(W=\int F d x=-(0.1) \int x d x=\frac{1}{2} m u^2-\frac{1}{2} m v^2\)

∴ final \(\mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2}(10 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)^2-0.1\left[\frac{x^2}{2}\right]_{20 \mathrm{~m}}^{30 \mathrm{~m}}\)

⇒ \(=500 \mathrm{~J}-\frac{0.1}{2}[900-400] \mathrm{J}\)

= 50Q J- 25 J

= 475 J.

Question 18. The force F acting on an object varies with the distance x as shown in the adjoining figure. The forceFisinnewtons and x are in metres. The work done by the force in moving the object from x = θ to x = 6m is

1. 9.0 J
2. 4.5 J
3. 13.5 J
4. 18.0 J

Answer: 3. 13.5 J

The work done by a variable force is given by

W = \(\int d W=\int F d x\) = area under the F-x graph.

From the given figure, the total work done is

W= area of the rectangle + area of the triangle

⇒ \((3 m)(3 N)+\frac{1}{2}(3 m)(3 N)\)

= 13.5J.

work energy and power neet questions

Question 19. The force F on a particle moving along a straight line varies with the distance d as shown in the following figure. The work done on the particle during its displacement of 12m is

1. 13 J
2. 18 J
3. 21 J
4. 30 J

Answer: 1. 13 J

From the given figure, the force from d = 0 to d = 3m is zero.

So, the work done is

⇒ \(W=(7-3) m \times(2 N)+\frac{1}{2}(12 m-7 m)(2 N)\)

= 8J + 5J

= 13J.

Question 20. A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude P₀. The instantaneous velocity of this car is proportional to

1. \(t^2 P_0\)
2. t^1/2
3. t ^-1/2
4. \(\frac{t}{\sqrt{m}}\)

Answer: 2. t^1/2

Instantaneous power is given by

⇒ \(P_0=F v=\left(m \frac{d v}{d t}\right) v\)

∴ P₀dt = mv dv.

Integrating, \(P_0 t=\frac{m v^2}{2}\)

∴ velocity = \(v=\sqrt{\frac{2 P_0 t}{m}}\)

∴ \(v \propto t^{1 / 2}\).

Question 21. A particle of mass m is driven by a machine that delivers a constant power P. If the particle starts from rest, the force on the particle at a time t is

1. \(\sqrt{\frac{2 m P}{t}}\)
2. \(\frac{1}{2} \sqrt{\frac{m P}{t}}\)
3. \(\sqrt{\frac{m P}{2 t}}\)
4. \(\sqrt{\frac{m P}{t}}\)

Answer: 3. \(\sqrt{\frac{m P}{2 t}}\)

Constant power = P = \(\frac{d W}{d t}\)

So, W = \(\int d W=\int P d t=P t\)

Now, work done = change in the KE = \(\frac{1}{2} m v^2-0=\frac{1}{2} m v^2\)

∴ \(P t=\frac{1}{2} m v^2 \text { or } v=\sqrt{\frac{2 P t}{m}}\)

Now, acceleration = a = \(\frac{d v}{d t}=\sqrt{\frac{2 P}{m}}\left(\frac{1}{2 \sqrt{t}}\right)\)

The force on the particle is

⇒ \(F=m a=m \cdot \sqrt{\frac{2 P}{m}}\left(\frac{1}{2 \sqrt{t}}\right)=\sqrt{\frac{m P}{2 t}}\)

work energy and power neet questions

Question 22. A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 x 10^-4J at the end of the second revolution after the beginning of the motion?

1. 0.2 ms^-2
2. 0.15 m s^-2
3. 0.1 m s^-2
4. 0.18 m s^-2

Answer: 3. 0.1 m s^-2

Given: mass = m = 10 g = 10 x 10^-3 kg,

radius = R = 6.4 cm = 6.4 x 10^-2 m,

final KE = 8x 10^-4J

and initial KE = 0.

By the work-energy theorem,

work done = change in the KE

=> maT. 2(2πR) = KE_f.

Hence, tangential acceleration = aT = \(\frac{\mathrm{KE}_{\mathrm{f}}}{4 \pi R m}\)

⇒ \(\frac{8 \times 10^{-4} \mathrm{~J}}{4 \times 3.14 \times 6.4 \times 10^{-2} \times 10^{-2} \mathrm{~kg} \mathrm{~m}}\)

= 0.099m s^-2

= 0.1m s^-2

Question 23. Consider a drop of rainwater of mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m s^-1 Take g equal to 10 m s^-2 The work done by the gravitational force and that by the resistive force of air are respectively

1. 1.25 J and -8.25 J
2. 100 J and 8.75 J
3. 10 J and -8.75 J
4. -10 J and -8.25 J

Answer: 3. 10 J and -8.75 J

The work done by the gravitational force is

⇒ \(W_{\text {grav }}=m g h=\left(1 \times 10^{-3} \mathrm{~kg}\right)\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(1000 \mathrm{~m})\)

= 10 J.

If W_fric = work done by the resistive force then

⇒ \(W_{\text {grav }}+W_{\text {fric }}=\text { change in the } \mathrm{KE}=\frac{1}{2} m v^2-0\)

⇒ \(10 \mathrm{~J}+W_{\text {fric }}=\frac{1}{2}\left(10^{-3} \mathrm{~kg}\right)\left(50 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\)

= 1.25 J.

Hence, \(W_{\text {fric }}=(1.25-10) \mathrm{J}\)

= -8.75 J.

Question 24. A body of mass 1 kg begins to move under the action of a time-dependent force \(\vec{F}=\left(2 t \hat{i}+3 t^2 \hat{j}\right) \mathrm{N}, \text { where } \hat{i} \text { and } \hat{j}\) are the unit vectors along the x- and y-axes. The power developed by the force at a time t will be

1. (2t³ + 3t⁴)W
2. (2t³ + 3t⁵)W
3. (2t² + 3t³)W
4. (2t² + 4t⁴)W

Answer: 2. (2t³ + 3t⁵)W

Given: force = \(F=\left(2 t \hat{i}+3 t^2 \hat{j}\right)\) N and mass = m =1 kg.

∴ acceleration = \(\vec{a}=\frac{\vec{F}}{m}=\frac{\left(2 t \hat{i}+3 t^2 \hat{j}\right) \mathrm{N}}{1 \mathrm{~kg}}\)

Hence, the velocity at the time t is

⇒ \(\vec{v}=\int_0^t \vec{a} d t=\int_0^t\left(2 t \hat{i}+3 t^2 \hat{j}\right) d t=\left(t^2 \hat{i}+t^3 \hat{j}\right) \mathrm{m} \mathrm{s}^{-1}\)

∴ the instantaneous power delivered at the time t will be

⇒ \(P=\vec{F} \cdot \vec{v}=\left(2 t \hat{i}+3 t^2 \hat{j}\right) \mathrm{N} \cdot\left(t^2 \hat{i}+t^3 \hat{j}\right) \mathrm{m} \mathrm{s}^{-1}=\left(2 t^3+3 t^5\right) \mathrm{W}\)

work energy and power neet questions

Question 25. What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

1. √gR
2. √2gR
3. √5gR
4. √3gR

Answer: 3. √5gR

⇒ At the highest point, the centripetal force (mv2/R) is provided by the weight (mg).

So,

⇒ \(\frac{m v^2}{R}=m g \text { and } \mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2} m g R\)

By the work-energy principle,

⇒ \(W_{\text {grav }}=\mathrm{KE}_{\mathrm{f}}-\mathrm{KE}_{\mathrm{i}}\)

or \(m g \cdot 2 R=\frac{1}{2} m v_{\mathrm{f}}^2-\frac{1}{2} m g R\)

∴ the required speed is \(v_{\mathrm{f}}=\sqrt{5 g R}\)

Question 26. The heart of a man pumps 5 liters of blood through the arteries per minute at a pressure of 150 mmHg. If the density of mercury be 13.6 x 10³ kg m^-3 and g = 10 m s^-2 then the power is

1. 1.50 W
2. 2.35 W
3. 3.0 W
4. 1.70 W

Answer: 3. 3.0 W

The flow rate of blood is \(5 \mathrm{~L} \mathrm{~m}^{-1}=\frac{5 \times 10^{-3} \mathrm{~m}^3}{60 \mathrm{~s}}\)

Blood pressure =p = 150mmHg = hpg

= (150 x 10^-3m)(13.6 x 10³kgm^-3)(10m s^-2)

= 20.4 x 10³Pa.

∴ the power of the heart is

⇒ \(P=\frac{W}{t}=\frac{p V}{t}=\frac{\left(20.4 \times 10^3 \mathrm{~Pa}\right)\left(5 \times 10^{-3} \mathrm{~m}^3\right)}{60 \mathrm{~s}}\)

= 1.70 W.

Question 27. A bullet is fired from a rifle which recoils after firing. The ratio of the kinetic energy of the rifle to that of the bullet is

1. One
2. Zero
3. Less Than One
4. More Than One

Answer: 3. Less Than One

Let M = mass of the gun, m = mass of die bullet, V = recoil speed of the gun, and v = speed of the bullet.

By the principle of conservation of linear momentum, MV = mv = p.

⇒ \(\mathrm{KE}_{\text {gun }}=\frac{p^2}{2 M} \text { and } \mathrm{KE}_{\text {bullet }}=\frac{p^2}{2 m}\)

Now,

∴ \(\frac{\mathrm{KE}_{\text {gun }}}{\mathrm{KE}_{\text {bullet }}}=\frac{m}{M}<1\)

Question 28. A body of mass 0.5 kg, moving at a speed of 1.5 m s^-1 on a smooth horizontal surface, collides with a light spring of force constant k = 5ON m^-1. The maximum compression of the spring will be

1. 1.5 m
2. 0.15 m
3. 0.5 m
4. 0.12 m

Answer: 2. 0.15 m

By the principle of conservation of energy,

KE of the body = PE in the spring

⇒ \(\frac{1}{2} m v^2=\frac{1}{2} k x^2\)

Here, the maximum compression of the spring is

⇒ \(x=\sqrt{\frac{m v^2}{k}}=\sqrt{\frac{(0.5)(1.5)^2}{50}} \mathrm{~m}\)

= 0.15m.

Question 29. A block of mass m moves up an inclined plane of inclination 0. If p is the coefficient of friction between the block and the inclined plane, the work done in moving the block through a distance s up along the plane is

1. mgs (sin θ + μ cos θ)
2. mgs sin θ
3. mgs(sm θ- μ cos θ)
4. μmgs cos θ

Answer: 1. mgs(sin 0 + μ cos 0)

Since the block is pulled up the incline, the applied force will overcome the components of the weight (mg sin 9) and the kinetic friction (mmg cos 0).

Hence, the work done by the pulling force is

W = Fs = (mg sin θ + μmg cos θ)s

= mgs(sin θ + μ cos θ).

Question 30. Amovingbullethits a solid target resting on a frictionless horizontal surface and gets embedded in the target. What is conserved in this process?

1. The momentum and the kinetic energy
2. The momentum alone
3. The kinetic energy alone
4. Neither the momentum nor the kinetic energy

Answer: 2. The momentum alone

In all types of collisions, it is only the momentum that is conserved.

Question 31. The potential energy of a particle in a force field is \(U=\frac{A}{r^2}-\frac{B}{r}\), where A and B are positive constants and r is the distance of the particle from the center of the field. For stable equilibrium, the distance of the particle is

1. \(\frac{B}{2 A}\)
2. \(\frac{B}{A}\)
3. \(\frac{2 A}{B}\)
4. \(\frac{A}{B}\)

Answer: 3. \(\frac{2 A}{B}\)

Given that PE = \(U(r)=\frac{A}{r^2}-\frac{B}{r}\)

For equilibrium,

⇒ \(\frac{d U}{d r}=0=-\frac{2 A}{r^3}+\frac{B}{r^2} \Rightarrow r=\frac{2 A}{B}\)

For stable equilibrium,

⇒ \(\frac{d^2 U}{d r^2}>0 \Rightarrow \frac{d^2 U}{d r^2}=\frac{6 A}{r^4}-\frac{2 B}{r^3}\)

Substituting r = \(\frac{2A}{B}\), we have

⇒ \(\frac{d^2 U}{d r^2}=6 A\left(\frac{B}{2 A}\right)^4-2 B\left(\frac{B}{2 A}\right)^3=\frac{B^4}{8 A^3}>0\)

Thus, for stable equilibrium,

⇒ \(r=\frac{2 A}{B}\)

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Question 32. A body projected vertically from the earth reaches a height equal to the earth’s radius before returning to the earth’s surface. When is the power delivered by the gravitational force of the earth the greatest?

1. At the highest position of the body
2. At the instant before the body hits the ground
3. Constant all through
4. At the instant after the body is projected

Answer: 2. At the instant before the body hits the ground

The instantaneous power delivered is P= Fv.

Just before hitting the ground, the velocity v is maximum, so the power delivered is also maximum.

Question 33. A body of mass 5 kg rests on a rough horizontal surface having a coefficient of friction of 0.2. The body is pulled through a distance of 10 m by a horizontal force of 25 N. The kinetic energy acquired by it is (assuming g = 10 m s^-2)

1. 200 J
2. 150 J
3. 100 J
4. 50 J

Answer: 2. 150 J

The work done by the applied force is

⇒ \(W_1=\vec{F} \cdot \vec{s}=F s \cos \theta\)

= (25 N)(10 m) cos θ°

= 250 J,

and the work done by friction is

⇒ \(W_2=-(\mu m g) s\)

= -(0.2)(5)(10)(10) J

= -100 j.

Net work done = change in the KE.

∴ KE acquired = W₁ + W₂

= 250 J- 100 J

= 150 J.

work energy and power neet questions

Question 34. A body is moving up an inclined plane of angle 0 with an initial kinetic energy E. The coefficient of friction between the plane and the body is p. The work done against the friction before the body comes to rest is

1. \(\frac{\mu \cos \theta}{E \sin \theta+\cos \theta}\)
2. \(\mu E \cos \theta\)
3. \(\frac{\mu E \cos \theta}{\mu \cos \theta+\sin \theta}\)
4. \(\frac{\mu E \cos \theta}{\mu \cos \theta-\sin \theta}\)

Answer: 3. \(\frac{\mu E \cos \theta}{\mu \cos \theta+\sin \theta}\)

By the work-energy theorem,

gain in KE = E = total work done

⇒ \(W_{\text {grav }}+W_{\text {fric }}=m g s \sin \theta+W_{\text {fric }}\)

∴ \(W_{\text {fric }}=E-m g s \sin \theta\)……(1)

Now, the upward acceleration along the rough plane is

a = g(sin0 + mcos 0).

Hence, u² = 2as= 3g (sinθ+ m cos θ)s

or, \(s=\frac{u^2}{2 g(\sin \theta+\mu \cos \theta)}\)

Substituting s in(1),

⇒ \(W_{\text {fric }}=E-\frac{m g \sin \theta \cdot u^2}{2 g(\sin \theta+\mu \cos \theta)}\)

⇒ \(E-\frac{E \sin \theta}{\sin \theta+\mu \cos \theta}=\frac{\mu E \cos \theta}{\mu \cos \theta+\sin \theta}\)

Question 35. A bullet is fired normally on an immovable wooden plank. It loses 25% of its momentum penetrating a thickness of 3.5 cm. The total thickness penetrated by the bullet is

1. 7 cm
2. 8 cm
3. 10 cm
4. 12 cm

Answer: 2. 8 cm

KE = \(\frac{p^2}{2 m}\)

Loss in momentum = \((25 \% \text { of } p)=\frac{p}{4}\)

∴ remaining momentum = \(\frac{3 p}{4}\)

∴ final \(\mathrm{KE}=\frac{\left(\frac{3 p}{4}\right)^2}{2 m}=\frac{9 p^2}{32 m}\)

According to the work-energy theorem,

⇒ \(F(3.5 \mathrm{~cm})=\Delta \mathrm{KE}=\frac{p^2}{2 m}-\frac{9 p^2}{32 m}=\frac{7 p^2}{32 m}\)

At the moment when the bullet finally comes to rest,

⇒ \(F(s)=\Delta K E=\frac{p^2}{2 m}-0=\frac{p^2}{2 m}\)

⇒ \(\frac{F(s)}{F(3.5 \mathrm{~cm})}=\frac{\frac{p^2}{2 m}}{\frac{7 p^2}{32 m}}=\frac{16}{7}\)

⇒ \(s=(3.5 \mathrm{~cm}) \frac{16}{7}\)

= 8 cm

work energy and power neet questions

Question 36. A uniform rod of mass m and length is held inclined at an angle of 60° with the vertical. What will be its potential energy in this position?

1. \(\frac{m g l}{4}\)
2. \(\frac{m g l}{3}\)
3. \(\frac{m g l}{2}\)
4. mgl

Answer: 1. \(\frac{m g l}{4}\)

PE of the uniform rod

= weight x height of the CM from the ground

⇒ \(m g\left(\frac{l}{2} \cos 60^{\circ}\right)\)

⇒ \(\frac{1}{4} m g l\)

Question 37. An engine pulls a car of mass 1500 kg on a level road at a constant speed of 18 km h^-1. If the frictional force is 1500N, what power does the engine generate?

1. 5.0kW
2. 7.5 kW
3. 10 kW
4. 12.5 kW

Answer: 2. 7.5 kW

Since the car moves at a constant velocity (c = 5ms ^-2), acceleration = 0.

Hence, F_net = 0.

The force F exerted by the engine must be equal and opposite to the frictional force (F = 1500 N).

∴ the power delivered by the engine is

P = Fv = (1500N)(5m s^-1)

= 7500 W

= 7.5kW

Question 38. In the preceding question, what extra power must the engine develop to maintain the same speed up along an inclined plane having a gradient of 1 in 10? (Take g = 10 m s^-2.)

1. 2.5 kW
2. 5.0 kW
3. 7.5 kW
4. 10 kW

Answer: 3. 7.5 kW

Extra power required = (mg sin θ)v

⇒ \((1500 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)\left(\frac{1}{10}\right)\left(5 \mathrm{~m} \mathrm{~s}^{-1}\right)\)

= 7500 W

= 7.5kW.

Question 39. Each of two identical cylindrical vessels, with their bases at the same level, contains a liquid of density p. The height of the liquid in one vessel is hx and that in the other is h₂. The area of either base is A. What is the work done by gravity in equalizing the levels when the vessels are interconnected?

1. \(A \rho g\left(h_1-h_2\right)^2\)
2. \(A \rho g\left(h_1+h_2\right)^2\)
3. \(A \rho g\left(\frac{h_1-h_2}{2}\right)^2\)
4. \(A \rho g\left(\frac{h_1+h_2}{2}\right)^2\)

Answer: 3. \(A \rho g\left(\frac{h_1-h_2}{2}\right)^2\)

The work done by gravity is equal to the change in the PE of the system.

The total initial PE of the system is

⇒ \(U_i=m_1 g\left(\frac{h_1}{2}\right)+m_2 g\left(\frac{h_2}{2}\right)\)

⇒ \(A h_1 \rho g\left(\frac{h_1}{2}\right)+A h_2 \rho g\left(\frac{h_2}{2}\right)=\frac{A \rho g}{2}\left(h_1^2+h_2^2\right)\)

When the vessels are interconnected, the height of the liquid in each

⇒ \(\frac{h_1+h_2}{2}\).

Hence, the final PE of the system is

⇒ \(U_{\mathrm{f}}=2\left[A\left(\frac{h_1+h_2}{2}\right) \rho g\left\{\frac{\left(h_1+h_2\right) / 2}{2}\right\}\right]=\frac{A \rho g}{4}\left(h_1+h_2\right)^2\)

∴ the change in the PE is

⇒ \(\Delta U=U_{\mathrm{f}}-U_{\mathrm{i}}=\frac{A \rho g}{4}\left(h_1+h_2\right)^2-\frac{A \rho g}{2}\left(h_1^2+h_2{ }^2\right)\)

⇒ \(\frac{A \rho g}{4}\left[\left(h_1+h_2\right)^2-2\left(h_1^2+h_2^2\right)\right]=-\frac{A \rho g}{4}\left(h_1-h_2\right)^2\)

∴ the work done by gravity is

⇒ \(W=-\Delta U=A \rho g\left(\frac{h_1-h_2}{2}\right)^2\)

Question 40. An electric pump on the ground floor of a building takes 10 minutes to fill a tank of volume 30 m3 with water. If the tank is 60 m above the ground and the efficiency of the engine is 30%, how much electric power is consumed by the pump in filling the tank? (Takeg = 10 m s^-2.)

1. 100 kW
2. 150 kW
3. 200 kW
4. 250 kW

Answer: 1. 100 kW

Output power = \(\frac{\text { work }}{\text { time }}=\frac{m g h}{t}=\frac{V \rho g h}{t}\)

⇒ \(\frac{\left(30 \mathrm{~m}^3\right)\left(10^3 \mathrm{~kg} \mathrm{~m}^{-3}\right)\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(60 \mathrm{~m})}{10 \times 60 \mathrm{~s}}\)

= 30 x 10³ W.

⇒ \(\text { Efficiency }=\frac{\text { output power }}{\text { input power }}\)

⇒ \(30 \%=\frac{30}{100}=\frac{30 \times 10^3 \mathrm{~W}}{\text { input power }}\)

⇒ input power = power consumed by the engine

⇒ \(\left(30 \times 10^3 \mathrm{~W}\right) \times \frac{100}{30}\)

= 100kW.

work energy and power neet questions

Question 41. A ball of mass m is thrown vertically upwards with a velocity of v. The height at which the kinetic energy of the ball will reduce to half its initial value is given by

1. \(\frac{v^2}{g}\)
2. \(\frac{v^2}{2 g}\)
3. \(\frac{v^2}{3 g}\)
4. \(\frac{v^2}{4 g}\)

Answer: 4. \(\frac{v^2}{4 g}\)

Initial KE = E = \(\frac{1}{2} m v^2\)

At the height h,KE = \(\frac{E}{2}=\frac{1}{4} m v^2\)

By the work-energy theorem,

⇒ \(W_{\text {grav }}=\Delta \mathrm{KEE}=\frac{E}{2}-E=-\frac{E}{2}\)

⇒ \(-m g h=-\frac{1}{4} m v^2 \Rightarrow h=\frac{v^2}{4 g}\)

Question 42. In a hydroelectric power station, the height of the dam is 10 m. How many kilograms of water must fall per second on the blades of a turbine so as to generate 1 MW of electrical power? (Take g = 10 m s^-2.)

1. 10³
2. 10⁴
3. 10⁵
4. 10⁶

Answer: 2. 10⁴

Power generated =1 MW = 10⁶ W = \(\frac{m g h}{t}\)

Given that h = 10m,g = 10m s^-2 and t = 1 s.

∴ \(m=\frac{\left(10^6 \mathrm{~W}\right)(1 \mathrm{~s})}{\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(10 \mathrm{~m})}=10^4 \mathrm{~kg} \mathrm{~s}^{-1}\)

Question 43. A metal ball of mass 2 kg, moving at a speed of 36 km h, undergoes a perfectly inelastic head-on collision with a stationary ball of mass 3 kg. The loss in its kinetic energy during the collision is

1. 40 J
2. 100 J
3. 140 J
4. 60 J

Answer: 4. 60 J

By the conservation of linear momentum,

⇒ \(m_1 u=\left(m_1+m_2\right) v \Rightarrow v=\frac{m_1 u}{m_1+m_2}\)

Initial KE = \(\frac{1}{2}(2 \mathrm{~kg})\left(36 \times \frac{5}{18} \mathrm{~m} \mathrm{~s}^{-1}\right)^2=100 \mathrm{~J}\)

and final KE = \(\frac{1}{2}\left(m_1+m_2\right) \frac{m_1^2 u^2}{\left(m_1+m_2\right)^2}=\frac{1}{2}\left(\frac{m_1^2 u^2}{m_1+m_2}\right)\)

⇒ \(\frac{\frac{1}{2}(2 \mathrm{~kg})^2\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)^2}{5 \mathrm{~kg}}=40 \mathrm{~J}\)

∴ loss in KE = 100j – 40J

= 60J.

Question 44. A ball is thrown vertically downwards from a height of 20 m with an initial velocity v₀. It collides with the ground, loses 50% of its energy during the collision, and rebounds to the same height. The initial velocity v₀ is equal to

1. 10 ms^-1
2. 14 ms^-1
3. 28 ms^-1
4. 20 ms^-1

Answer: 4. 20 ms^-1

Let the velocity of the ball just before it strikes the ground be v.

⇒ \(v^2=v_0^2+2 g h\)

⇒ \(\frac{1}{2} m v^2=\frac{1}{2} m v_0^2+m g h\)

⇒ \(E=E_0+m g h\)

The remaining energy E’ after the 50% loss will be

⇒ \(E^{\prime}=\frac{E}{2}=\frac{1}{2}\left(E_0+m g h\right)\)

∴ \(\frac{1}{2}\left(\frac{1}{2} m v_0^2+m g h\right)=m g h\) [ … at h, velocity = 0]

⇒ \(v_0=\sqrt{2 g h}=\sqrt{2\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(20 \mathrm{~m})}\)

= 20m s^-1.

Question 45. A bullet of mass 10 g, moving horizontally at a velocity of 400m s^-1, strikes a wooden block of mass 2 kg, which is suspended by a light, inextensible string of length 5 m. As a result, the center of gravity of the block is found to rise vertically through 10 cm. The speed of the bullet after it emerges horizontally from the block will be

1. 80 ms^-1
2. 100 ms^-1
3. 120 ms^-1
4. 160 ms^-1

Answer: 3. 120 ms^-1

Given that m = 10 g, u = 400m s^-1, and M = 2 kg.

If V₁ = velocity of the block just after the collision,

⇒ \(\frac{1}{2} M v_1^2=M g h\)

⇒ \(\dot{v}_1=\sqrt{2 g h}=\sqrt{2(10)\left(10 \times 10^{-2}\right)} \mathrm{ms}^{-1}\)

⇒ \(\sqrt{2} \mathrm{~ms}^{-1}\)

By the conservation of linear momentum,

mu =Mv₁ + mv

⇒ \(\left(10 \times 10^{-3} \mathrm{~kg}\right)\left(400 \mathrm{~m} \mathrm{~s}^{-1}\right)=(2 \mathrm{~kg})\left(\sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\right)+\left(10 \times 10^{-3} \mathrm{~kg}\right) v\)

⇒ \(v=\frac{4-2 \sqrt{2}}{0.01} \mathrm{~m} \mathrm{~s}^{-1}=117.15 \mathrm{~m} \mathrm{~s}^{-1} \approx 120 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 46. A block of mass m, moving at a speed v on a frictionless horizontal surface, collides elastically with another block of the same mass m, initially at rest. After the collision, the first block moves at an angle 0 to the initial direction and has a speed of  The speed of the second block after the collision is 3

1. \(\frac{\sqrt{3}}{2} v\)
2. \(\frac{3}{4} v\)
3. \(\frac{3}{\sqrt{2}} v\)
4. \(\frac{2 \sqrt{2}}{3} v\)

Answer: 4. \(\frac{2 \sqrt{2}}{3} v\)

Let vx be the speed of the second block. In an elastic collision, the kinetic energy is conserved.

So,

⇒ \(\frac{1}{2} m v^2+0=\frac{1}{2} m\left(\frac{v}{3}\right)^2+\frac{1}{2} m v_1^2\)

⇒ \(v_1^2=v^2-\frac{v^2}{9}=\frac{8}{9} v\)

⇒ \(v_1=\frac{2 \sqrt{2}}{3} v\)

work energy and power neet questions

Question 47. A body of mass 4m is lying in the xy-plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass m, fly off perpendicular to each other with equal speeds of v. The total kinetic energy generated due to the explosion is

1. mv²
2. \(\frac{3}{2} m v^2\)
3. 2mv²
4. 4 muv²

Answer: 2. \(\frac{3}{2} m v^2\)

The resultant of the momentum vectors of A and B is √2mv.

Before the explosion, the momentum is zero.

So, for the final momentum to be zero, we have from the adjoining figure,

⇒ \(2 m v^{\prime}=\sqrt{2} m v \Rightarrow v^{\prime}=\frac{v}{\sqrt{2}}\)

∴ the KE generated due to the explosion is

⇒ \(\frac{1}{2} m v^2+\frac{1}{2} m v^2+\frac{1}{2}(2 m)\left(\frac{v}{\sqrt{2}}\right)^2\)

⇒ \(\frac{3}{2} m v^2\)

Question 48. A mass m, moving horizontally (along the *-axis) at a velocity v, collides with sand sticks to a mass of 3m, moving vertically upwards (along die y-axis) at a velocity 2v. The final velocity of the combination is

1. \(\frac{1}{3} v \hat{i}+\frac{2}{3} v \hat{j}\)
2. \(\frac{2}{3} v \hat{i}+\frac{1}{3} v \hat{j}\)
3. \(\frac{3}{2} v \hat{i}+\frac{1}{4} v \hat{j}\)
4. \(\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}\)

Answer: 4. \(\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}\)

The initial momentum of the mass m is \(\vec{p}_1=m v \hat{i}\), and that of the mass

⇒ \(3 m \text { is } \vec{p}_2=3 m(2 v) \hat{j}\)

∴ the total initial momentum is

⇒ \(\vec{p}_1=\vec{p}_1+\vec{p}_2=m v \hat{i}+6 m v \hat{j}\)

Finally, when the masses stick and move together, let the final momentum be \(\vec{p}_{\mathrm{f}}=4 m \vec{v}_0\)

Conserving the momentum, we have

⇒ \(4 m \vec{v}_0=m v \hat{i}+6 m v \hat{j} \Rightarrow \vec{v}_0=\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}\)

Question 49. A body of mass 2 kg has an initial velocity \(\vec{v}_i=(\hat{i}+\hat{j}) \mathrm{m} \mathrm{s}^{-1}\). After a collision with another body, its velocity becomes \(\vec{v}_{\mathrm{f}}=(5 \hat{i}+6 \hat{j}+\hat{k}) \mathrm{m} \mathrm{s}^{-1}\). If the time of impact is 0.02 s, the average force of impact on the body is

1. \(100(4 \hat{i}+5 \hat{j}-\hat{k}) \mathrm{N}\)
2. \(100(4 \hat{i}+5 \hat{j}+\hat{k}) \mathrm{N}\)
3. \(50(4 \hat{i}-5 \hat{j}-\hat{k}) \mathrm{N}\)
4. \(50(4 \hat{i}+5 \hat{j}+\hat{k}) \mathrm{N}\)

Answer: 2. \(100(4 \hat{i}+5 \hat{j}+\hat{k}) \mathrm{N}\)

Initial momentum = \(\vec{p}_{\mathrm{i}}=(2 \mathrm{~kg})(\hat{i}+\hat{j}) \mathrm{m} \mathrm{s}^{-1}\)

and final momentum = \(\vec{p}_{\mathrm{f}}=(2 \mathrm{~kg})(5 \hat{i}+6 \hat{j}+\hat{k}) \mathrm{m} \mathrm{s}^{-1}\)

∴ changeinmomentum = \(\Delta \vec{p}=\vec{p}_{\mathrm{f}}-\vec{p}_{\mathrm{i}}\)

⇒ \((2 \mathrm{~kg})(4 \hat{i}+5 \hat{j}+\hat{k}) \mathrm{m} \mathrm{s}^{-1}\)

∴ average force = \(\vec{F}_{\mathrm{av}}=\frac{\Delta \vec{p}}{\Delta t}=\frac{(2 \mathrm{~kg})(4 \hat{i}+5 \hat{j}+\hat{k}) \mathrm{m} \mathrm{s}^{-1}}{0.02 \mathrm{~s}}\)

⇒ \(100(4 \hat{i}+5 \hat{j}+\hat{k}) \mathrm{N}\)

work energy and power neet questions

Question 50. A ball of mass m moving with a velocity v undergoes an oblique elastic collision with another ball of the same mass at rest. After the collision, if the two balls move at equal speeds, the angle between their directions of motion will be

1. 60°
2. 30°
3. 90°
4. 120°

Answer: 3. 90°

Let \(\vec{p}_1 \text { and } \vec{p}_2\) be the momenta of the two balls after the elastic collision.

Hence, conserving the momentum, we have

⇒ \(\vec{p}=\overrightarrow{p_1}+\overrightarrow{p_2} \text {, where } \vec{p}\) = initial momentum.

Taking the self-dot-product,

⇒ \(\vec{p} \cdot \vec{p}=\left(\overrightarrow{p_1}+\overrightarrow{p_2}\right) \cdot\left(\overrightarrow{p_1}+\overrightarrow{p_2}\right)\)

or \(p^2=p_1^2+p_2^2+2 \vec{p}_1 \cdot \vec{p}_2\)

or \(\frac{p^2}{2 m}=\frac{p_1^2}{2 m}+\frac{p_2^2}{2 m}+\frac{2 \vec{p}_1 \cdot \overrightarrow{p_2}}{2 m}\) ……(1)

⇒ In an elastic collision, the KE (=p²/2m) is conserved.

∴ \(E=E_1+E_2 \text { or } \frac{p^2}{2 m}=\frac{p_1^2}{2 m}+\frac{p_2^2}{2 m}\) …(2)

From (1) and(2),

⇒ \(\overrightarrow{p_1} \cdot \overrightarrow{p_2}=0 \text {, i.e., } \overrightarrow{p_1} \text { is perpendicular to } \overrightarrow{p_2}\)

So, the angle between the velocities of the two balls is 90°.

Question 51. A body of mass 2 kg moving at a velocity \((\hat{i}+2 \hat{j}-3 \hat{k}) \mathrm{m} \mathrm{s}^{-1}\) collides with another body of mass 3 kg moving at a velocity \((2 \hat{i}+\hat{j}+\hat{k}) \mathrm{m} \mathrm{s}^{-1}\). If they stick together, the velocity of the composite body will be

1. \(\frac{1}{5}(8 \hat{i}+7 \hat{j}-3 \hat{k}) \mathrm{m} \mathrm{s}^{-1}\)
2. \(\frac{1}{5}(-4 \hat{i}+\hat{j}-3 \hat{k}) \mathrm{m} \mathrm{s}^{-1}\)
3. \(\frac{1}{5}(8 \hat{i}+\hat{j}-\hat{k}) \mathrm{m} \mathrm{s}^{-1}\)
4. \(\frac{1}{5}(-4 \hat{i}+7 \hat{j}-3 \hat{k}) \mathrm{m} \mathrm{s}^{-1}\)

Answer: 1. \(\frac{1}{5}(8 \hat{i}+7 \hat{j}-3 \hat{k}) \mathrm{m} \mathrm{s}^{-1}\)

Conserving the momentum,

⇒ \((2 \mathrm{~kg})(\hat{i}+2 \hat{j}-3 \hat{k}) \mathrm{m} \mathrm{s}^{-1}+(3 \mathrm{~kg})(2 \hat{i}+\hat{j}+\hat{k}) \mathrm{m} \mathrm{s}^{-1}\)

⇒ \((2 \mathrm{~kg}+3 \mathrm{~kg}) \vec{v}\)

or, \((5 \mathrm{~kg}) \vec{v}=(8 \hat{i}+7 \hat{j}-3 \hat{k}) \mathrm{kg} \mathrm{m} \mathrm{s}^{-1}\)

Hence, the velocity of the composite body is

⇒ \(\vec{v}=\frac{1}{5}(8 \hat{i}+7 \hat{j}-3 \hat{k}) \mathrm{m} \mathrm{s}^{-1}\)

work energy and power neet questions

Question 52. A steel ball falls from a height on a floor for which the coefficient of restitution is e. The height attained by the ball after two rebounds is

1. eH
2. e²H
3. e³H
4. e⁴H

Answer: 4. e⁴H

We know that the coefficient of restitution (e) is expressed by the relation velocity of separation

= e(velodty of approach).

The initial speed (u) when the ball hits the ground is given by

⇒ \(u^2=2 g H \quad \Rightarrow \quad u=\sqrt{2 g H}\)

The first recoil speed is u₁ = eu and the second recoil speed is

⇒ \(u_2=e u_1=e(e u)=e^2 u=e^2 \sqrt{2} g H .\)

If H² is the maximum height,

⇒ \(u_2^2=2 g H_2 \Rightarrow 2 g H_2=e^4(2 g H)\)

=> H² = e⁴H.

work energy and power neet questions

Question 53. A body of mass m, moving at a constant speed, undergoes an elastic collision with a body of mass m₂, initially at rest. The ratio of the kinetic energy of the first body after the collision to that before the collision is

1. \(\left(\frac{m_1-m_2}{m_1+m_2}\right)^2\)
2. \(\left(\frac{m_1+m_2}{m_1-m_2}\right)^2\)
3. \(\left(\frac{2 m_1}{m_1+m_2}\right)^2\)
4. \(\left(\frac{2 m_1}{m_1-m_2}\right)^2\)

Answer: 1. \(\left(\frac{m_1-m_2}{m_1+m_2}\right)^2\)

Conserving the momentum,

⇒ \(m_1 u=m_1 v_1+m_2 v_2\)….(1)

where u is the initial speed of the ball before the collision, and v₁ and v₂ are the speeds of the balls after the collision.

For an elastic collision, the coefficient of restitution = e = 1.

∴ \(\left(v_2-v_1\right)=1 \times u \Rightarrow v_2=u+v_1\)

Substituting for v₂ in (1), we get

⇒ \(m_1 u=m_1 v_1+m_2\left(u+v_1\right)\)

=> (m₁– m₂)u = (m₁ + m₂)v₁

⇒ \(v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right) u\)

⇒ \(\frac{\mathrm{KE}_{\text {after }}}{\mathrm{KE}_{\text {before }}}=\frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_1 u^2}=\left(\frac{v_1}{u}\right)^2=\left(\frac{m_1-m_2}{m_1+m_2}\right)^2\)

Question 54. A radioactive nucleus of mass number A, initially at rest, emits an a-particle with a speed v. What will be the recoil speed of the daughter nucleus?

1. \(\frac{2 v}{A-4}\)
2. \(\frac{2 v}{A+4}\)
3. ⇒ \(\frac{4 v}{A-4}\)
4. \(\frac{4 v}{A+4}\)

Answer: 3. \(\frac{4 v}{A-4}\)

The scheme of decay is shown in the adjoining figure. After the decay,

themomentaofthe fragments are \(4 m v \hat{i} \text { and }(A-4) m v^{\prime}(-\hat{i})\) respectively.

Conserving the momentum,

⇒ \(4 m v \hat{i}-(A-4) m v^{\prime} \hat{i}=0 \Rightarrow 4 v=(A-4) v^{\prime}\)

⇒ \(v^{\prime}=\frac{4 v}{A-4}\)

Question 55. Consider a system comprising a light string, a light spring, and a block of mass M suspended from a rigid support, as shown in the adjacent figure. The spring constant is k and the block is released when the spring

1. \(\frac{M g}{2}\)
2. Mg
3. 3Mg
4. 2Mg

Answer: 4. 2Mg

The initial of the block is zero, and finally, at the maximum extension (x) of the file spring, it comes to momentary rest.

Thus, by the work-energy theorem,

⇒ \(W_{\mathrm{grav}}+W_{\mathrm{sp}}=\Delta \mathrm{KE}=0\)

⇒ \(M g x+\left(-\frac{1}{2} k x^2\right)=0 \Rightarrow x=\frac{2 M g}{k}\)

∴ the maximum tension in the spring is

⇒ \(F_{\max }=k x_{\max }=k\left(\frac{2 M g}{k}\right)\)

= 2Mg.

Question 56. In the given figure, the spring has its natural length, the blocks are at rest, and there is no friction anywhere. If the spring constant is k= 20Nm^-1, the maximum extension in the spring will be

1. \(\frac{10}{3} \mathrm{~cm}\)
2. \(\frac{20}{3} \mathrm{~cm}\)
3. \(\frac{40}{3} \mathrm{~cm}\)
4. \(\frac{19}{3} \mathrm{~cm}\)

Answer: 1. \(\frac{10}{3} \mathrm{~cm}\)

When the extension in the spring is maximum (= x0), the blocks have the same velocity and the same acceleration (= a).

The system can be considered as connected bodies, and we apply Newton’s second law from the given free-body diagram.

For A, F-kx₀ = m₁a…(1)

For B, kx₀ = m₂a….(2)

Adding (1) and (2),

⇒ \(a=\frac{F}{m_1+m_2}\)….(3)

Substituting a from (3) in (2), the maximum extension becomes

⇒ \(x_0=\frac{m_2 a}{k}=\frac{m_2 F}{k\left(m_1+m_2\right)}\)

Substituting the given values

⇒ \(x_0=\frac{(1 \mathrm{~kg})(1 \mathrm{~N})}{\left(20 \mathrm{~N} \mathrm{~m}^{-1}\right)\left(\frac{3}{2} \mathrm{~kg}\right)}=\frac{10}{3} \mathrm{~cm}\)

Question 57. Find the maximum force if the potential energy is \(U=\frac{a}{r^2}-\frac{b}{r}\) (Given that a = 2 and b = 4.)

1. \(-\frac{16}{27} \mathrm{~N}\)
2. \(-\frac{32}{27} \mathrm{~N}\)
3. \(+\frac{32}{27} \mathrm{~N}\)
4. \(+\frac{16}{27} \mathrm{~N}\)

Answer: 1. \(-\frac{16}{27} \mathrm{~N}\)

By definition, \(F=-\frac{d U}{d r} \Rightarrow F=\frac{2 a}{r^3}-\frac{b}{r^2}\)

For F to be maximum,

⇒ \(\frac{d F}{d r}=0 \Rightarrow-\frac{6 a}{r^4}+\frac{2 b}{r^3}=0\)

⇒ \(r=\frac{3 a}{b}=3\left(\frac{2}{4}\right)=\frac{3}{2}\)

∴ \(F_{\max }=\frac{2 a}{r^3}-\frac{b}{r^2}\)

⇒ \(\frac{2(2)}{\left(\frac{3}{2}\right)^3} \mathrm{~N}-\frac{4}{\left(\frac{3}{2}\right)^2} \mathrm{~N}\)

⇒ \(\frac{32}{27} \mathrm{~N}-\frac{16}{9} \mathrm{~N}\)

⇒ \(-\frac{16}{27} \mathrm{~N}\)

Question 58. Initially when the spring is relaxed, having its natural length, block A of mass 0.25 kg is released. Find the maximum force exerted by the system on the floor.

1. 20N
2. 15N
3. 30N
4. 25N

Answer: 4. 25N

By the work-energy theorem,

⇒ \(W_{\text {grav }}+W_{\text {sp }}=\Delta \mathrm{KE}=0\)

⇒ \(m_{\mathrm{A}} g x_0-\frac{1}{2} k x_0^2=0\), where x0= maximum compression

⇒ \(x_0=\frac{2 m_A g}{k}\)

∴ the maximum force on the floor is

⇒ \(F_{\max }=m_{\mathrm{B}} g+k x_0=m_{\mathrm{B}} g+2 m_{\mathrm{A}} g\)

⇒ \(\left(m_{\mathrm{B}}+2 m_{\mathrm{A}}\right) \mathrm{g}=(2 \mathrm{~kg}+0.5 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)\)

= 25N.

Question 59. A force F = 20x + 10y acts on a particle in the y-direction, where F is in newtons and y is in meters. The work done by this force to move the particle from y = 0 to y = 1 m is

1. 30 J
2. 5 J
3. 25 J
4. 20 J

Answer: 3. 25 J

The total work done is

⇒ \(W=\int d W=\int \vec{F} \cdot d \vec{y}=\int_0^1m(20+10 y) d y\)

⇒ \(\left[20 y+5 y^2\right]_0^{1 m}\)

= 25J

Question 60. A constant force acts on a 2-kg object so that its position is given as a function of time by x = 3t²-5. What is the work done by this force in the first five seconds?

1. 850 J
2. 875 J
3. 900 J
4. 950 J

Answer: 3. 900 J

Since x = 3t2 – 5, the instantaneous velocity is

⇒ \(v=\frac{d x}{d t}=6 t\)

and the acceleration is

⇒ \(a=\frac{d v}{d t}=6 \mathrm{~m} \mathrm{~s}^{-2}=\text { constant }\)

∴ force = F = ma = (2kg)(6m s^-2)

= 12N.

The distance covered in 5 s is

⇒ \(x=u t+\frac{1}{2} a t^2=0 \cdot t+\frac{1}{2}\left(6 \mathrm{~m} \mathrm{~s}^{-2}\right)\left(25 \mathrm{~s}^2\right)\)

= 75m.

∴ work done =Fx = (12N)(75m) = 900 J

Question 61. A particle undergoes a displacement under the action of a \(\vec{r}=4 \hat{i}\) constant force \(\vec{F}=3 \hat{i}-12 \hat{j}\);. If the initial kinetic energy of the particle is 3 J, the kinetic energy at the end of the displacement will be

1. 9 J
2. 15 J
3. 12 J
4. 10 J

Answer: 2. 15 J

Work done = \(\vec{F} \cdot \vec{r}=(3 \hat{i}-12 \hat{j}) \mathrm{N} \cdot 4 \hat{i} \mathrm{~m}\)

=12 J.

∴ change in the KE = KE_f – KE_i = work done.

∴ \(\mathrm{KE}_{\mathrm{f}}=3 \mathrm{~J}+12 \mathrm{~J}=15 \mathrm{~J}\)

Question 62. Ablockofmassm, lying on a smooth horizontal surface, is attached to a light spring of spring constant k. The other end of the spring is fixed, as shown in the adjoining figure. The block is initially at rest in its equilibrium position. If the block is now pulled with a constant force F, the maximum speed of the block will be

1. \(\frac{\pi F}{\sqrt{m k}}\)
2. \(\frac{F}{\sqrt{m k}}\)
3. \(\frac{2 F}{\sqrt{m k}}\)
4. \(\frac{F}{\pi \sqrt{m k}}\)

Answer: 2. \(\frac{F}{\sqrt{m k}}\)

As the force produces elongation (x) in the spring, a retarding force (-kx) develops and we have

F_net= F-kx = ma.

When the velocity is maximum, the acceleration \(\left(\frac{d v}{d t}\right)\) is zero.

Thus,

⇒ \(F-k x=0 \Rightarrow x=\frac{F}{k}\)

According to the work-energy theorem,

⇒ \(W_{\text {fric }}+W_{\text {sp }}=\mathrm{KE}_{\max }=\frac{1}{2} m v_{\max }^2\)

⇒ \(F x-\frac{1}{2} k x^2=\frac{1}{2} m v_{\max }^2\)……(1)

Substituting x = \(\frac{F}{k}\) in (1) the condition for the maximum velocity,

⇒ \(F\left(\frac{F}{k}\right)-\frac{1}{2} k\left(\frac{F}{k}\right)^2=\frac{1}{2} m v_{\max }^2\)

⇒ \(\frac{F^2}{2 k}=\frac{1}{2} m v_{\max }^2 \Rightarrow v_{\max }=\frac{F}{\sqrt{m k}}\)

Question 63. A block of mass m is kept on a platform that starts from rest with a constant acceleration \(\frac{g}{2}\) upwards, as shown in the adjoining figure. The work done by the normal reaction on the block in a time t will be

1. \(\frac{m g^2 t^2}{8}\)
2. \(\frac{3 m g^2 t^2}{8}\)
3. \(-\frac{m g^2 t^2}{8}\)
4. Zero

Answer: 2. \(\frac{3 m g^2 t^2}{8}\)

The height reached by the block in the time t is

⇒ \(h=\frac{1}{2}\left(\frac{g}{2}\right) t^2=\frac{g t^2}{4}\)

and the velocity at that time is

⇒ \(v=\left(\frac{g}{2}\right) t=\frac{g t}{2}\)

The work done by gravity is

⇒ \(W_{\mathrm{grav}}=-(m g) h=-m g\left(\frac{g t^2}{4}\right)\)

Change in the

⇒ \(\mathrm{KE}=\frac{1}{2} m\left(\frac{g t}{2}\right)^2=\frac{1}{2}\left(\frac{m g^2 t^2}{4}\right)\)

From the work-energy theorem,

⇒ \(\mathrm{W}_{\mathrm{grav}}+W_{O V}=\frac{1}{2} m v^2\)

⇒ \(-\frac{m g^2 t^2}{4}+W_{\delta V}=\frac{1}{2} m\left(\frac{g^2 t^2}{4}\right)\)

∴ the work done by the normal reaction is

⇒ \(W N =\frac{m g^2 t^2}{4}+\frac{m g^2 t^2}{8}=\frac{3}{8} m g^2 t^2\)

Question 64. Throe blocks A, B, and C are placed on a smooth horizontal surface, as shown in the given figure. A and B have equal masses of m, while C has a mass of M. Block A is given an initial speed of v towards B, due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically. If in the whole process, \(\frac{5}{6}\) of its initial kinetic energy is lost, the value of the ratio \(\frac{M}{m}\) is

1. 4
2. 2
3. 3
4. 5

Answer: 1. 4

Since momentum is conserved during each impact, p = mu = (M + 2m) v₀, where v₀ is the final velocity of the combined system.

Initial kinetic energy of A is \(\frac{p^2}{2 m}\) and final kinetic energy of the combined system is

⇒ \(\frac{1}{6}\left(\frac{p^2}{2 m}\right), \text { since } \frac{5}{6} \text { part in loss }\)

⇒ \(\frac{1}{6}\left(\frac{p^2}{2 m}\right)=\frac{p^2}{2(M+2 m)}\)

or, 6m = (M+2m)

Hence \(\frac{M}{m}=4 .\)

Question 65. A gun exerts a time-dependent force on a bullet given by

F = [100- (0.5 x 10⁵)t] N.

If the bullet emerges with a speed of 400 m s^-1, find the impulse delivered to the bullet till the force F is reduced to zero.

1. 0.1 Ns
2. 0.3Ns
3. 0.5Ns
4. 0.4Ns

Answer: 1. 0.1 Ns

When F is reduced to zero at a time t, we have

⇒ \(t=\frac{100}{(0.5) 10^5} \mathrm{~s}=\frac{2}{1000} \mathrm{~s}\)

The impulse delivered to the bullet is

⇒ \(I=\int_0^t F d t=\left[100 t-\left(0.5 \times 10^5\right) \frac{t^2}{2}\right]_0^{(2 / 1000)}\)

⇒ \(100\left(\frac{2}{1000}\right)-\frac{1}{2} \times \frac{10^5}{2}\left(\frac{2}{10^3}\right)^2\)

=(0.2- 0.1)N s

= 0.1N s.

Question 66. A block A of mass 4m, moving towards block B of mass 2m at rest, undergoes an elastic head-on collision. The fraction of energy lost by the colliding body A is

1. \(\frac{4}{9}\)
2. \(\frac{5}{9}\)
3. \(\frac{8}{9}\)
4. \(\frac{2}{9}\)

Answer: 3. \(\frac{8}{9}\)

Let v₁ and v₂ be the velocities of the blocks A and B after the collision.

Conserving the momentum,

4mu₁ + 2m(0) = 4mv₁ +2mv₂

=> 2u₁= 2v₁+v₂…..(1)

For an elastic collision, e = 1.

Thus,

v₂-v₁ = e(u₁-u₂)

=> v₂-v₁ = u₁…..(2)

Subtracting(2) from(1),

⇒ \(3 v_1=u_1 \Rightarrow v_1=\frac{u_1}{3}\)

The fractional loss of energy the block A is

⇒ \(\frac{\Delta \mathrm{KE}_{\mathrm{A}}}{\mathrm{KE}_{\mathrm{A}}}=\frac{\frac{1}{2} m_{\mathrm{A}} u_1^2-\frac{1}{2} m_{\mathrm{A}} v_1^2}{\frac{1}{2} m_{\mathrm{A}} u_1^2}\)

⇒ \(1-\left(\frac{v_1}{u_1}\right)^2=1-\left(\frac{1}{3}\right)^2\)

⇒ \(\frac{8}{9}\)

Question 67. A block rests on a rough horizontal surface. A horizontal force that increases linearly with time (t) starts acting on the block at t = 0. Which of the following velocity-time graphs will be correct?

Answer: 2.

Given that \(F=m \frac{d v}{d t}=\alpha t \Rightarrow d v=\frac{\alpha}{m}, t d t\)

Integrating, \(v=\frac{\alpha}{2 m} \cdot t^2\)

Due to its roughness, the block starts moving after some time when

F≥f_k.

Question 68. A particle of mass m is moving at a speed 2v and collides with a particle of mass 2m, moving at a speed v in the same direction. After the collision, the first particle is stopped, while the second particle splits up into two smaller particles, each of mass m, which move at 45° with respect to the original direction. The speed of each moving particle will be

1. \(\frac{v}{2 \sqrt{2}}\)
2. 2 √2v
3. \(\frac{v}{\sqrt{2}}\)
4. √2v

Answer: 2. 2 √2v

The situations before and after the collision are shown in the figure below.

Conserving the momentum along the direction of motion,

m.2v + 2m.v = mv’cos45°+ mv’cos45°

⇒ \(4 m v=2 m v^{\prime}\left(\frac{1}{\sqrt{2}}\right)\)

⇒ \(v^{\prime}=\frac{4}{\sqrt{2}} v=2 \sqrt{2} v\)

Question 69. A body of mass m, moving at a velocity \(\hat{u i}\), collides elastically with a stationary body of mass 3m. If the velocity of the lighter body after the collision is \(\hat{v j}\) then this velocity has the magnitude After the collision

1. \(v=\frac{u}{2}\)
2. \(v=\frac{u}{\sqrt{3}}\)
3. \(v=\frac{u}{\sqrt{2}}\)
4. \(v=\frac{4}{3}\)

Answer: 3. \(v=\frac{u}{\sqrt{2}}\)

Conserving the KE, we have

⇒ \(\frac{1}{2} m u^2=\frac{1}{2} m v^2+\frac{1}{2}(3 m) v^{\prime 2}\)

or, \(u^2-v^2=3 v^2\)

Conserving the momentum, we have along the

x-axis: mu = 3mv’cosθ

and along the y-axis: 0 = mv-3mv’sin θ.

∴ \(u^2+v^2=9 v^{\prime 2}\)

From (1) and (2)

⇒ \(u^2+v^2=3\left(u^2-v^2\right) \Rightarrow v=\frac{u}{\sqrt{2}}\)

Question 70. Five blocks are placed on a smooth, horizontal, plane surface along the same line, as shown. The first block (of mass m) is given a velocity v so that the blocks successively undergo a perfectly inelastic collision. The percentage loss in the kinetic energy after the last collision is

1. 88.5%
2. 93.75%
3. 90.2%
4. 85.5%

Answer: 3. 90.2%

The momentum (p) is conserved during the collision.

However, due to the perfectly inelastic collision, the mass has changed from m to 16m.

Initial KE = \(\frac{p^2}{2 m} \text { and final } \mathrm{KE}=\frac{p^2}{2(16 m)}\)

∴ AKE = changee in the KE = \(\)

∴ % loss in KE = \(\frac{p^2}{2 m}-\frac{p^2}{2(16 m)}\)

Question 71. The force-displacement graph of a particle moving along the x-axis is shown in the given figure. The work done by the force from x = 0 to x = 30m is

1. 4550 J
2. 5250 J
3. 4250 J
4. 7500 J

Answer: 2. 5250 J

The area under the force-displacement graph gives the work done.

Hence,

W = area of the rectangle + area of the trapezium

= (200N)(15 m) + \(\frac{1}{2}\) (200N + 100 N)(15m)

= 3000 J + 2250 J

= 5250 J

Question 72. A block of mass m starts slipping down an inclined plane from the topmost point B and finally comes to rest at the lowermost point A. If BC = 2AC and the friction coefficient of the rough part ACis A m = k tan 6 then the value of k is

1. 0.5
2. 0.35
3. 0.75
4. 3.0

Answer: 4. 3.0

Let AC = x and BC = 2x.

∴ AB = 3x.

Work done by gravity = \(W_{\mathrm{grav}}=m g h=m g(3 x \sin \theta)\)

Work done by friction = \(W_{\text {fric }}=-f x=-(\mu m g \cos \theta) x\)

From the work-energy theorem,

⇒ \(W_{\text {grav }}+W_{\text {fric }}=\Delta K E=K_f-K_1=0\)

=> (mg)(3x) sin θ = (μmg cosθ)x

=> 3 tan θ = m = k tan 0 [ given ]

=> k=3.

Question 73. A 100-g bullet moving horizontally at 20m s^-1 strikes a stationary block of mass 1.9 kg and comes to rest inside. The kinetic energy of the block just before hitting the ground is

1. 11 J
2. 21 J
3. 24 J
4. 32 J

Answer: 2. 21 J

Conserving the linear momentum,

(100 g)(20m s^-1) = (100 g + 1.9 kg)v

=> v = 1m s^-1.

From the work-energy principle,

Wgrav = changein the KE

⇒ \((2 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)(1 \mathrm{~m})=\mathrm{KE}_{\mathrm{f}}-\mathrm{KE}_{\mathrm{i}}\)

=\(\mathrm{KE}_{\mathrm{f}}-\frac{1}{2}(2 \mathrm{~kg})\left(1 \mathrm{~m} \mathrm{~s}^{-1}\right)^2\)

Hence, KE_f = 20J +1J

= 21J.

Question 74. A block is pushed up along a rough inclined plane with a speed V₀, as shown in the given figure. After some time, it is again back to the starting point with a reduced Find the coefficient of speed \(\frac{v_0}{2}\) fiction (μ).

1. 0.15
2. 0.45
3. 0.35
4. 0.75

Answer: 3. 0.35

Applying the work-energy theorem,

⇒ \(W_{\text {grav }}+W_{\text {fric }}=\Delta K E=K_f-K_i\)

The work done by gravity is zero, and the work done by friction is

⇒ \(W_{\text {fric }}=-\mu \delta N(2 s)=-\left(\mu m g \cos 30^{\circ}\right)(2 s)\)

The change in KE is

⇒ \(\frac{1}{2} m\left(\frac{v_0}{2}\right)^2-\frac{1}{2} m v_0^2=-\frac{1}{2}\left(\frac{3}{4} m v_0^2\right)\)

∴ \(\frac{1}{2} \cdot \frac{3}{4} m v_0^2=(\mu m g \sqrt{3}) s\)

⇒ \(v_0^2=\frac{8}{\sqrt{3}} \mu g s\)…..(1)

For demotion down the plane,

⇒ \(\left(\frac{v_0}{2}\right)^2=0+2 g(\sin \theta-\mu \cos \theta) s\)

⇒ \(v_0^2=8 g\left(\frac{1}{2}-\frac{\sqrt{3}}{2} \mu\right) s\) …..(2)

Equating (1) and(2),

⇒ \(4 g s(1-\sqrt{3} \mu)=\frac{8}{\sqrt{3}} \mu g s\)

⇒ \(\mu\left(\sqrt{3}+\frac{2}{\sqrt{3}}\right)=1\)

⇒ \(\mu=\frac{\sqrt{3}}{5}\)

= 0.35

Question 75. A block of mass 2 kg at rest is driven by a constant power of 1.0 W. The distance covered by the block in 6 s is

1. 2√6 m
2. 4√6 m
3. 3√3 m
4. 4√3 m

Answer: 2. 4√6 m

Instantaneous power P = Fv = \(\left(m \frac{d v}{d t}\right) v\)

∴ \(\frac{P}{m} d t=v d v\)

Integrating, \(\left(\frac{P}{m}\right) t=\frac{y^2}{2}\)

Hence, velocity = \(v=\frac{d s}{d t}=\sqrt{\frac{2 P}{m}} t^{1 / 2}\)

Integrating again,

⇒ \(s=\sqrt{\frac{2 P}{m}}\left(\int t^{1 / 2} d t\right)=\sqrt{\frac{2 P}{m}}\left(\frac{2}{3} t^{3 / 2}\right)=\frac{2}{3} \sqrt{\frac{2 \times 1}{2}}\left(6^{3 / 2}\right)=4 \sqrt{6} \mathrm{~m}\)