Responses to Questions

Question 1.

Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.

• Suma is swimming in a pond.
  • Work is done as she exerts a force against the water to move.
• A donkey is carrying a load on its back.
  • No work is done as the load is carried horizontally with no displacement in the direction of the force acting on it (gravity).
• A wind-mill is lifting water from a well.
  • Work is done as the wind-mill is exerting force to lift water against gravity.
• A green plant is carrying out photosynthesis.
  • No work is done in the physical sense as there is no displacement of the plant.
• An engine is pulling a train.
  • Work is done as the engine exerts a force along the direction of the movement of the train.
• Food grains are getting dried in the sun.
  • No work is done as there is no displacement in the direction of the force acting on the grains (heat energy dispersing).
• A sailboat is moving due to wind energy.
  • Work is done as the wind exerts a force on the sails of the boat resulting in movement.

Question 2.

An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?

• Answer: The work done by gravity is zero because the overall vertical displacement of the object is zero (initial and final heights are the same).

Question 3.

A battery lights a bulb. Describe the energy changes involved in the process.

• Answer: The chemical energy stored in the battery is converted into electrical energy, which then transforms into light energy and thermal energy in the bulb.

Question 4.

Certain force acting on a 20 kg mass changes its velocity from 5 m s–1 to 2 m s–1. Calculate the work done by the force.

• Answer:
  Initial kinetic energy (KE*initial) = 0.5 * 20 kg _ (5 m/s)^2 = 250 J
  Final kinetic energy (KE_final) = 0.5 _ 20 kg _ (2 m/s)^2 = 40 J
  Work done = KE_final - KE_initial = 40 J - 250 J = -210 J (negative work done).

Question 5.

A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.

• Answer: The work done by gravitational force is zero because there is no vertical displacement; the movement is horizontal.

Question 6.

The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

• Answer: No, it does not violate the law of conservation of energy because the potential energy lost by the object is converted into kinetic energy, thus maintaining the total energy constant.

Question 7.

What are the various energy transformations that occur when you are riding a bicycle?

• Answer: When riding a bicycle, mechanical energy (from pedaling) is converted into kinetic energy (motion of the bicycle), and some energy is lost as thermal energy due to friction.

Question 8.

Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?

• Answer: No transfer of energy takes place to the rock because it does not move. The energy expended is converted into heat energy due to friction and internal energy of the muscles used in the push.

Question 9.

A certain household has consumed 250 units of energy during a month. How much energy is this in joules?

• Answer: 1 unit (kilowatt-hour) = 3.6 × 10^6 J
  Therefore, 250 units = 250 × 3.6 × 10^6 J = 900000000 J or 9 × 10^8 J.

Question 10.

An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.

• Answer:
  Potential Energy (PE) = mgh = 40 kg × 9.8 m/s² × 5 m = 1960 J
  At halfway down, the height is 2.5 m.
  Kinetic Energy (KE) = Initial PE - PE at halfway = 1960 J - (40 kg × 9.8 m/s² × 2.5 m) = 1960 J - 980 J = 980 J.

Question 11.

What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

• Answer: The work done by gravity on a satellite in circular motion is zero because there is no displacement in the direction of the gravitational force; the satellite is constantly changing direction but not altitude.

Question 12.

Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.

• Answer: No, displacement requires a force to initiate and maintain movement. Without force, an object at rest remains at rest, and an object in motion continues in motion only if no external forces act upon it.

Question 13.

A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.

• Answer: No mechanical work has been done in the physics sense because the bundle is not displaced in the direction of the force (upwards) despite the person experiencing fatigue due to exertion.

Question 14.

An electric heater is rated 1500 W. How much energy does it use in 10 hours?

• Answer:
  Energy = Power × Time = 1500 W × 10 hours = 1500 W × 36000 s = 54000000 J or 54 × 10^6 J.

Question 15.

Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?

• Answer: When the bob is raised, potential energy increases, and when released, it converts to kinetic energy. The bob eventually comes to rest due to energy loss from friction and air resistance, converting kinetic energy into thermal energy. This is not a violation of conservation of energy; the total energy remains conserved in the system including all forms.

Question 16.

An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?

• Answer: The work done must equal the initial kinetic energy to bring it to rest. Work = KE = 0.5 _ m _ v².

Question 17.

Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?

• Answer:
  Convert 60 km/h to m/s: 60 km/h = (60 × 1000 m)/(3600 s) = 16.67 m/s
  KE = 0.5 _ m _ v² = 0.5 _ 1500 kg _ (16.67 m/s)² ≈ 208,350 J.
  Thus, work done = -208,350 J (negative since it's stopping).

Question 18.

In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.

• Answer:

• If the force is in the same direction as displacement, work done is positive.

• If the force is in the opposite direction of displacement, work done is negative.

• If there is no displacement in the direction of the applied force, work done is zero.

Question 19.

Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?

• Answer: Yes, I agree. An object can have zero acceleration if the net force acting on it is zero, as in a state of equilibrium where forces are balanced.

Question 20.

Find the energy in joules consumed in 10 hours by four devices of power 500 W each.

• Answer:
  Total power = 4 devices × 500 W = 2000 W
  Energy = Power × Time = 2000 W × 10 hours = 2000 W × 36000 s = 72000000 J or 72 × 10^6 J.

Question 21.

A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

• Answer: When the object reaches the ground and stops, its kinetic energy is converted to other forms, primarily thermal energy due to the impact and deformation, or it is dissipated as sound energy.