## Chapter 4 Work And Energy Long Answer Type Question And Answers

**Question 1. A bullet of mass 50 g travelling horizontally with a speed of 200 ms ^{-1} strikes a glass pave 4 mm thick and falls dead, on emerging from it. How much work is done by the bullet against the opposing force of the glass pave?**

**Answer.**

**Given:**

A bullet of mass 50 g travelling horizontally with a speed of 200 ms^{-1} strikes a glass pave 4 mm thick and falls dead, on emerging from it.

Initial K.E. of the bullet = \(\frac{1}{2} m u^2\)

Final K.E. of the bullet = \(\frac{1}{2} m v^2\)

Loss in K.E. of the bullet = Initial K.E. – Final K.E.

= \(\frac{1}{2} m u^2-\frac{1}{2} m v^2\)

= \(\frac{1}{2} m\left(u^2-v^2\right)\)

By Work – Energy theorem,

Work done by a body = Loss in its K.E.

W = \(\frac{1}{2} m\left(u^2-v^2\right)\)

= \(\frac{1}{2} \times \frac{50}{1000}\left(200^2-0^2\right)\)

= \(\frac{1}{2} \times \frac{5}{100} \times 200 \times 200\)

= 1000J

= 1 x 10^{3} J

**Question 2. If a heavy truck (mass = M) and a light car (mass m which is less than M) possess same amount of momentum, then which one possess greater kinetic energy? Explain.**

**Answer.**

We know that momentum ‘p’ of a body of mass ‘m’ moving with velocity v is given by

p = mv

Squaring on both sides,

p^{2} = (mv)^{2} = m^{2}v^{2}

p^{2} = m⋅mv

= \(2 m\left(\frac{1}{2} m v^2\right)\)

p^{2} = 2mE_{k}

So, K.E., \(E_\kappa=\frac{p^2}{2 m}\)

If p is same, i.e., constant, then

\(E_K=\frac{C}{m}\)i.e., E_{k} is inversely proportional to mass.

It means, a body with smaller mass possess more K.E.

∴ Car will possess more K.E. than the truck.

**Question 3. A body fall from height H. If t _{1} is time taken for covering the first half height and t_{2} be the time taken for second half. Which of these relations is true for t_{1} and t_{2}.**

**t**_{1}> t_{2}**t**_{1}< t_{2}**t**_{1}= t_{2}**Depends on the mass of the body**

**Answer.**

**Given:**

A body fall from height H. If t_{1} is time taken for covering the first half height and t_{2} be the time taken for second half.

Let H be the height, then

First Half

\(\frac{H}{2}=\left(\frac{1}{2}\right) g t_1^2\) (1)

Or \(\left(\frac{1}{2}\right) g t_1^2=\frac{H}{2}\)

Also v = gt_{1}

Second Half

\(\frac{H}{2}=v t_2+\left(\frac{1}{2}\right) g t_2^2\)or \(\left(\frac{H}{2}\right)=g t_1 t_2+\left(\frac{1}{2}\right) g t_2^2\)

or \(\left(\frac{1}{2}\right) g t_2^2=\left(\frac{H}{2}\right)-g t_1 t_2\) (2)

t_{2}^{2} + 2t_{1}t_{2} − t_{1}^{2} = 0

or \(t_2=\frac{\left[-2 t_1+\sqrt{\left(4 t_1^2+4 t_1^2\right)}\right]}{2}\)

or \(t_2=\frac{\left[-2 t_1+2 t_1 \sqrt{2}\right]}{2}\)

t_{2} = 0.4 t_{1}

so, t_{1} > t_{2}

Hence a is correct.

**Question 4. A delivery boy wishes to launch a 2.0 kg package up an inclined plane with sufficient speed to reach the top of the incline. The plane is 3 m long and is inclined at 20. the coefficient of friction between the package and the inclined plane is .40.What minimum initial K.E. must the boy supply to the package given as sin20 = .342 cos20 = .940?**

**Answer.**

**Given:**

A delivery boy wishes to launch a 2.0 kg package up an inclined plane with sufficient speed to reach the top of the incline.

The plane is 3 m long and is inclined at 20. the coefficient of friction between the package and the inclined plane is .40.

If the package travels the entire length S of the incline, the frictional force will perform work − μNS where μ is the coefficient of friction and N is normal reaction.

Let ‘h’ be the height of the incline plane then the gravitational potential energy of the package will increase by mgh.

Now let us assume ‘v’ be the speed given to the package so as to reach the top

Then kinetic energy at the initial point = \(\frac{1}{2} m v^2\)

Now applying work energy theorem

K.E_{f} – K.E_{i} = Work done by the gravitational force + Work done by the frictional force

Now since K.E_{f} = 0

Also Work done by the gravitational force

= –(change in gravitational potential energy)

= −mgh

Therefore

\(\frac{1}{2} m v^2\) = mgh – μNS

or \(\frac{1}{2} m v^2\) = mgh + μN

Now S = 3

N = mgcosθ

h = S sinθ

Substituting all the values

\(\frac{1}{2} m v^2\) = 42.2 J

**Question 5. Deduce a formula for K.E of a body.**

**Answer.**

**A formula for K.E of a body:**

Consider a body of mass ‘m’ starts moving from rest, with uniform velocity ‘u’. After a time interval ‘t’ its velocity becomes v.

If initial velocity of the body is u or vi = 0, final velocity vf = v and the displacement of body is ‘S’. Then

First of all we will find the acceleration of the body.

Using equation of motion

2aS = v_{f}^{2} − v_{i}^{2}

Substituting the above mentioned values

2aS = v^{2} − 0

a = \(\frac{v^2}{2 S}\)

Now force is given by

F = ma

Substituting the value of acceleration

F = \(m\left(\frac{v^2}{2 S}\right)\)

As we know that,

Work done = F.S

Substituting the value of F

Work done = \(\left(\frac{m v^2}{2 S}\right)(S)\)

Work done = \(\frac{m v^2}{2}\)

or Work done = \(\frac{1}{2} m v^2\)

Since the work done in motion is called ‘Kinetic Energy’

i.e., K.E. = Work done

or \(\frac{1}{2} m v^2\)

**Question 6. A body of mass 1.0 kg initially at rest is moved by a horizontal force of 0.5 N on a smooth frictionless table. Calculate the work done by the force in 10 s and show that this is equal to the change in K.E. of the body.**

**Answer.**

**Given:**

A body of mass 1.0 kg initially at rest is moved by a horizontal force of 0.5 N on a smooth frictionless table.

Here, m = 1.0 kg, u = 0, F = 0.5 N, t = 10s

a = \(\frac{F}{m}=\frac{0.5}{1.0}=0.5 \mathrm{~ms}^{-2}\)

From S = \(u t+\frac{1}{2} a t^2\)

S = \(0+\frac{1}{2} \times 0.5(10)^2=2.5 \mathrm{~m}\)

Work done = F × S

= 0.5 × 25

= 12.5 J

From v = u + at = 0 + 0.5 × 10

= 5 ms–1

Change in K.E. = \(\frac{1}{2} m\left(v^2-u^2\right)\)

= \(\frac{1}{2} \times 1.0\left(5^2-0\right)\)

= 12.5 J