Work and Energy Worksheet6

What kind of energy is possessed by the following?
A stretched rubber band.

What are the quantities on which the amount of work done depends? How are they related to work?

Is it possible that a force is acting on a body but still the work done is zero? Explain giving one example.

A boy throws a rubber ball vertically upwards. What type of work, positive or negative, is done :
(a) by the force applied by the boy ?
(b) by the gravitational force of earth ?

Write the formula for work done on a body when the body moves at an angle to the direction of force. Give the meaning of each symbol used.

How does the kinetic energy of a moving body depend on its (i) speed, and (ii) mass?

Give one example each in which a force does (a) positive work (b) negative work, and (c) zero work.

A ball of mass 200 g falls from a height of 5 metres. What is its kinetic energy when it just reaches the ground? (g = 9.8 m/s^{2}).

Find the momentum of a body of mass 100 g having a kinetic energy of 20 J.

Two objects having equal masses are moving with uniform velocities of 2 m/s and 6 m/s respectively Calculate the ratio of their kinetic energies.

A body of 2 kg falls from rest. What will be its kinetic energy during the fall at the end of 2 s?
(Assume g = 10 m/s^{2})

On a level road, a cyclist applies brakes to slow down from a speed of 10 m/s to 5 m/s. If the mass of the cyclist and the scooter be 150 kg, calculate the work done by the brakes. (Neglect air resistance and friction)

A man drops a 10 kg rock from the top of a 5 m ladder. What is its speed just before it hits the ground? What is its kinetic energy when it reaches the ground ? (g = 10 m/s^{2})

Calculate the work done by the brakes of a car of mass 1000 kg when its speed is reduced from 20 m/s to 10 m/s?

A body of mass 100 kg is lifted up by 10 m. Find :
(i) the amount of work done.
(ii) potential energy of the body at that height (value of g = 10 m/s^{2}).

A boy weighing 50 kg climbs up a vertical height of 100 m. Calculate the amount of work done by him. How much potential energy does he gain ? (g = 9.8 m/s^{2}).

To what height should a box of mass 150 kg be lifted, so that its potential energy may become 7350 joules? (g = 9.8 m/s^{2}).

A body of mass 2 kg is thrown vertically upwards with an initial velocity of 20 m/s. What will be its potential energy at the end of 2 s? (Assume g = 10 m/s^{2}).

How much work is done when a force of 1 N moves a body through a distance of 1 m in its own direction?

A car is being driven by a force of 2.5 × 10^{10} N. Traveling at a constant speed of 5 m/s, it takes 2 minutes to reach a certain place. Calculate the work done.
Answer:

Potential energy

Work done in moving a body is equal to the product of force and the displacement of the body in the direction of force.
Work = Force ´ Displacement in the direction of force
Or Work = Force ´ Displacement
Or W = F ´ s

Yes it is possible if displacement is zero then work done will be zero.
Example : Moving a box from its position and then moving it back to its initial position. In this case displacement is zero so work done will be zero.

(a) Positive (b) Negative

W = f × s cos α
F = force applied
α = angle between the direction of force and direction of motion
s = distance moved

Kinetic energy = (1/2)mv^{2}

Explanation#:
a. If we kick a ball then the ball moves in the direction of force so the work done is positive.
b. A ball moving on the ground slows down and finally stops. Because a force due to friction acts on the ball and it acts in a direction opposite to the direction of motion so the work done is negative.
c. W = f × s cos α
F = force applied
α = angle between the direction of force and direction of motion
s = distance moved
When α is 90 degree work done will be zero .
The force of gravity of earth that acts on the satellite at right angle to the direction of motion of satellite so the work done is zero.

9.8 J = mgh

2 kg.m/s
Kinetic energy = (1/2)mv^{2} = 0.1 × V^{2}/2 = 20
V = 20 m/s
Momentum p = mv
0.01 × 20 = 0.2 kgm/sec

1 : 9
Using Kinetic energy = (1/2)mv^{2}

400 J
Kinetic energy = (1/2)mv^{2}

5625 J

10 m/s ;500 J

150 KJ

(i) 10,000 J (ii) 10,000 J

4.9 × 10^{4} J; 4.9 × 10^{4} J

5 m
PE = mgh
7350 = 150 × 9.8 × h
H = 5 m

400 J

1 J

1,5 × 10^{13} J