## Chapter 4 Work And Energy Short Answer Type Questions

**Question 1. A force of 20 N acts on a body to displace it through 10 m on a level road. Calculate the work done if force and displacement make an angle 60 degree with each other.**

**Answer. **

**Given:**

A force of 20 N acts on a body to displace it through 10 m on a level road.

Here force F = 20 N, Displacement, S = 10 m,

θ = 60 degrees

Now Work done, W = FS cos θ

= 20 × 10 × cos 60

= 20 × 10 × 0.5 = 100 J

**Question 2. Define work. Write the formula of work when force and displacement do not act in same direction.**

**Answer.**

**Work:**

Work is the product of force and displacement.

When force and displacement do not act in same direction, the formula for work is given by,

W = FS cos θ

Where θ is the angle between force and displacement.

**Question 3. Mention the difference between potential energy and kinetic energy. A sliding box of mass 1 kg slows down from 10 m/s to 4 m/s. Calculate the change in kinetic energy.**

**Answer.**

**Given:**

The difference between potential energy and kinetic energy. A sliding box of mass 1 kg slows down from 10 m/s to 4 m/s.

Potential energy of a body is the energy possessed by a body due to its position while kinetic energy of a body is the energy possessed due to velocity of the body.

Change in kinetic energy,

ΔK = 1/2m(v^{2} – u^{2})

ΔK = (42 – 10^{2})/2

= -84/2 = -42 J

**Question 4. Which physical quantity has Nm as its SI unit? Two different bodies of masses in the ratio of 1 : 2 are moving with same speed. What is the ratio of their kinetic energy?**

**Answer.**

**Given:**

Two different bodies of masses in the ratio of 1 : 2 are moving with same speed

Work has the SI unit Nm.

Ratio of Kinetic energies

\(\frac{K_1}{K_2}=\frac{\left(\frac{1}{2} m_1 v_1^2\right)}{\left(\frac{1}{2} m_2 v_2^2\right)}\) \(\frac{K_1}{K_2}=\left(\frac{m_1}{m_2}\right)\left(\frac{v_1^2}{v_2^2}\right)\) \(\frac{K_1}{K_2}=\frac{1}{2} \times 1 \quad\left(\text { since } v_1=v_2 \& \frac{m_1}{m_2}=\frac{1}{2}\right)\)So \(\frac{K_1}{K_2}\) = 0.5

**Question 5. In lifting a body to 10 m, 800 J of work was done. Calculate its weight.**

**Answer.**

**Given:**

In lifting a body to 10 m, 800 J of work was done.

Work done W = mgh

800 = mg × 10

mg = 800/10 = 80 N

Thus, weight of the body is 80 N.

**Question 6. How much work is done in lifting a book weighing 800 g through a height of 1.5 m? (Take g = 10 ms ^{-2})**

**Answer.**

We know that,

W = Force × displacement

⇒ W = mg × S

⇒ W = mgh (∵ S = h)

⇒ W = 0.8 kg × 10 ms^{-2} × 1.5 m

⇒ W = 12 J

**Question 7. When can work done by a force be zero? Give examples.**

**Answer.**

We know that, in general work done is given by

W = FS cos θ

Where θ is the angle between direction of force and displacement.

W = 0 = FS cos θ

⇒ either S = 0

or cos θ = 0

i.e., θ = 90°

So work done by a force is zero.

When either displacement is zero or angle between directions of force and displacement is zero.

Examples

1. When a force is applied to a body and it doesn’t move

Example: a wall.

2. Displacement is in horizontal direction and force in vertical direction. So, work done by gravity on a bucket, which a man is carrying along horizontal ground.

**Question 8. How much work is done in pulling a trolley through 6 m, as shown in the below figure?**

**Answer.**

We know that,

W = FS cos θ

= 100 N × 6 m × cos(60°)

= 100 × 6 × 0.5 Nm

= 300 J

**Question 9. You are standing on roof top of a tower of height 25 m. What is the potential energy of a ball of mass 150 g kept on the ground? (Take g = 10 ms ^{-2})**

**Answer.**

In this case, potential energy will be negative since the body is below the reference level which is on the roof top.

P.E. = mgh

= 0.150 kg × 10 ms^{-2} × (−25 m)

= −37.5 J

Note: This numerical tells us that potential energy of a body can be negative.

Example: Potential energy of coal in a coalmine is negative with respect to the observer on ground.

**Question 10. If the velocity of a body is doubled, how will its kinetic energy change? Compare new kinetic energy with the old one.**

**Answer.**

Consider a body of mass ‘m’ moving with a velocity ‘v1’.

Then, its

\(\text { K.E. }=\frac{1}{2} m v_1^2\)∴ \(E_1=\frac{1}{2} m v_1^2\) (1)

Now, its velocity is doubled.

So, v_{2 }= 2v_{1}

∴ Its new kinetic energy

\(E_2=\frac{1}{2} m v_2^2\)= \(\frac{1}{2} m\left(2 v_1^2\right)\)

= \(4 \cdot \frac{1}{2} m v_1^2\)

= 4 E_{1} form (1)

Thus, its kinetic energy becomes four times.

**Question 11. A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle. If the initial speed (in m/sec) of a particle is zero, than what is the speed (in m/sec) after 5 s?**

**Answer.**

**Given:**

A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle.

We know that

\(\text { Power }=\frac{d W}{d t}\)W = 0.5 × 5 = 2.5 = K.E._{f} − K.E._{i}

Since the initial speed of the particle is zero, the initial kinetic energy will also be zero.

⇒ \(2.5=\frac{M}{2}\left(v_f^2-v_i^2\right)\)

⇒ \(2.5=\frac{M}{2}\left(v_f^2-0\right)\)

⇒ \(v_f^2=2.5 \times \frac{2}{0.2}=25\)

⇒ v_{f} = 5

**Question 12. Deduce the formula of P.E. of a body.**

**Answer.**

**P.E. of a body:**

Consider an object of mass, m. Let it be raised through a height h from the ground. A force is required to do this. The minimum force required to raise the object is equal to the weight of the object mg.

The object gains energy equal to the work done on it. Let the work done on the object against gravity be W. That is, work done,

W = Force × Displacement

= mg × h

= mgh

Since work done on the object is equal to mgh, the energy equal to mgh units is gained by the object. This is the potential energy (EP) of the object.

E_{p} = mgh

**Question 13. When a rubber is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx ^{2}, where a and b are constants. What is the work done in stretching the un-stretched rubber b and by L?**

**Answer.**

**Given:**

When a rubber is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx^{2}, where a and b are constants

We have

F = ax + bx^{2}

W = Fdx

The work done by stretching the un-stretched rubber b and by L is

\(W=\int_0^L\left(a x+b x^2\right) d x\)= \(\left[a\left(\frac{x^2}{2}\right)+b\left(\frac{x^3}{3}\right)\right]_0^L\)

= \(a\left(\frac{L^2}{2}\right)+b\left(\frac{L^3}{3}\right)\)