Work and Energy Worksheet-6

Work and Energy Worksheet-6

 

  1. What kind of energy is possessed by the following?

A stretched rubber band.

 

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

 

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

 

  1. 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 ?

 

  1. 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.

 

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

 

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

 

  1. 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/s2).

 

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

 

  1. 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.

 

  1. 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/s2)

 

  1. 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)

 

  1. 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/s2)

 

  1. 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?

 

  1. 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/s2).

 

  1. 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/s2).

 

  1. 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/s2).

 

  1. 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/s2).

 

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

 

  1. A car is being driven by a force of 2.5 × 1010 N. Traveling at a constant speed of 5 m/s, it takes 2 minutes to reach a certain place. Calculate the work done.

 

Answer:

  1. Potential energy

 

  1. 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

 

  1. 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.

 

  1. (a) Positive (b) Negative

 

  1. W = f × s cos α

F = force applied

α = angle between the direction of force and direction of motion

s = distance moved

 

  1. Kinetic energy = (1/2)mv2

 

  1. 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.

 

  1. 9.8 J = mgh

 

  1. 2 kg.m/s

Kinetic energy = (1/2)mv2 = 0.1 × V2/2 = 20

V = 20 m/s

Momentum p = mv

0.01 × 20 = 0.2 kgm/sec

 

  1. 1 : 9

Using Kinetic energy = (1/2)mv2

 

  1. 400 J

Kinetic energy = (1/2)mv2

 

  1. 5625 J

 

  1. 10 m/s ;500 J

 

  1. 150 KJ

 

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

 

  1. 4.9 × 104 J; 4.9 × 104 J

 

  1. 5 m

PE = mgh

7350 = 150 × 9.8 × h

H = 5 m

 

  1. 400 J

 

  1. 1 J

 

  1. 1,5 × 1013 J