Work, Energy, and Power — Long Answer Questions (Class 9 Physics)
Medium Level (Application & Explanation)
Q1. Define work in physics. Give three real-life examples where no scientific work is done and explain why.
Answer:
Work in physics is done when a force causes an object to undergo displacement in the direction of the force. Mathematically, for constant force along the displacement, Work = Force × Displacement. The SI unit of work is the joule (J).
Examples where no scientific work is done:
Holding a heavy box stationary over your head: Although your muscles exert force and you feel tired, the box has no displacement, so no work is done on the box.
Pushing a wall that does not move: Your push applies force but the wall does not move, so displacement = 0 and no work is done.
Standing or sitting while reading a book: You may expend biological energy, but the book and your body do not undergo displacement in the direction of an applied external force, so no work is done in the physics sense.
Remember: effort ≠ work in physics; there must be displacement in the direction of the applied force.
Q2. Derive the expression for work done when lifting an object vertically through a height h. If a 10 kg object is lifted 2 m, calculate the work done against gravity (take g = 9.8 m/s²).
Answer:
When lifting an object vertically at constant speed, the applied upward force equals the gravitational force (mg) in magnitude. Displacement is vertical and equal to h. So work done against gravity: W = Force × Displacement = mg × h. This work increases the object’s gravitational potential energy.
For the given values: mass m = 10 kg, height h = 2 m, gravitational acceleration g = 9.8 m/s².
W = mg h = 10 × 9.8 × 2 = 196 J.
So 196 joules of work is done against gravity.
Note: If you lift the object faster or slower at constant speed, the work done against gravity remains the same; time does not appear in this expression.
Q3. Explain why work done by friction is considered negative. Give an everyday example and describe the energy change.
Answer:
Frictional force acts opposite to the direction of motion. When a force and displacement are in opposite directions, the scalar product (force × displacement) gives a negative value, so work done by friction is negative. Negative work means friction removes mechanical energy from the moving object, usually converting it into thermal energy (heat).
Example: Sliding a book across a table. You push the book forward; friction acts backward. The work done by friction is negative, so the book’s kinetic energy decreases and the table and book warm up slightly.
In energy terms: mechanical energy (kinetic) of the book is transformed into internal energy (heat) because of friction. This explains why objects eventually stop if no external force keeps them moving.
Q4. Define power. Two students lift identical 5 kg bags to a shelf 2 m high. Student A takes 4 s, Student B takes 2 s. Compare the power developed by each student. (g = 9.8 m/s²)
Answer:
Power is the rate at which work is done or energy is transferred. Mathematically, Power = Work / Time. The SI unit is watt (W) where 1 W = 1 J/s.
Work done in lifting one bag: W = mgh = 5 × 9.8 × 2 = 98 J.
Student A (time = 4 s): P_A = 98 / 4 = 24.5 W.
Student B (time = 2 s): P_B = 98 / 2 = 49 W.
Comparison: Student B develops twice the power of Student A because B does the same work in half the time. This shows that doing the same task faster requires greater power.
Q5. Distinguish between energy and work. Provide examples of kinetic and potential energy with short explanations.
Answer:
Energy is the capacity to do work. Work is the transfer of energy from one object or system to another. In simple terms: when work is done, energy is transferred or converted.
Kinetic energy (KE) is the energy of motion. For an object of mass m moving at speed v, KE = ½ m v². Example: a moving bicycle has kinetic energy which could do work if it hits a stationary object.
Potential energy (PE) is energy stored due to position or configuration. For gravitational potential energy near Earth’s surface, PE = m g h. Example: water stored in an overhead tank has gravitational potential energy which can be converted to kinetic energy when released.
Key point: Units of both are joules (J). Energy types change (KE ↔ PE), but the total energy is conserved in ideal situations without non-conservative forces.
High Complexity (Analytical & Scenario-Based)
Q6. A child pushes a toy car with a horizontal force of 8 N along a straight line for 4 m. There is a frictional force of 3 N opposing the motion. Calculate: (a) work done by the pushing force, (b) work done by friction, (c) net work on the car. Explain the physical meaning of net work.
Answer:
(a) Work by pushing force: W_push = F × s = 8 N × 4 m = 32 J (positive because force and displacement same direction).
(b) Work by friction: W_fric = (−3 N) × 4 m = −12 J (negative because friction opposes motion).
(c) Net work: W_net = W_push + W_fric = 32 J − 12 J = 20 J.
Physical meaning: Net work on the car equals the change in its kinetic energy (work-energy theorem). So the toy car’s kinetic energy increases by 20 J as a result of these forces. The positive net work indicates the car speeds up or gains kinetic energy. The frictional negative work reduced the amount of energy available to increase speed, and some energy converted into heat due to friction.
Q7. A person holds a bucket of water and walks across a flat floor at steady speed for 10 m. Discuss whether work is done on the bucket by the person, by gravity, and by the normal force. Explain energy changes in the system.
Answer:
When walking at steady speed on a flat floor, the bucket moves horizontally. Consider forces on the bucket: the person’s upward force balances weight, gravity acts downward, and the normal force from the person’s hand is upward; horizontal forces produce motion but if the person only supports weight and moves horizontally at constant speed, the net horizontal external force may be negligible.
Work by the person (vertical support): The vertical force the person applies does no net work related to vertical displacement because vertical displacement is zero; if the hand moves slightly up and down there can be small internal work, but overall no work done against gravity for horizontal motion.
Work by gravity and normal force: Both act perpendicular to the horizontal displacement, so their work is zero (since force × displacement × cos 90° = 0).
Energy changes: The person uses biological energy (internal chemical energy) to walk; this energy is partly used to overcome internal muscle inefficiencies and to overcome friction with the ground, producing heat. But from the physics point of view, no work is done on the bucket by these vertical forces during purely horizontal steady motion. The bucket’s kinetic energy remains approximately constant.
Q8. A spring of force constant k = 200 N/m is compressed by 0.10 m. Calculate the work done in compressing the spring and explain how this work is stored and can be recovered. (Use appropriate formula and explain energy conversion.)
Answer:
For a linear spring, the force varies with displacement: F = k x. The work done in compressing or stretching a spring from x = 0 to x = x₀ is the area under the force-displacement curve, given by W = ½ k x₀².
Here, k = 200 N/m, x₀ = 0.10 m, so W = ½ × 200 × (0.10)² = 0.5 × 200 × 0.01 = 1.0 J.
This 1.0 J of work is stored in the spring as elastic potential energy. If the spring is released, this stored energy is converted into kinetic energy of the attached object (and possibly other forms if non-ideal). In an ideal, frictionless system, the energy can be completely recovered as kinetic energy. In real systems some energy is lost as heat or sound, but the main idea is that mechanical work done to deform the spring is stored and later converted back when the spring returns to its original shape.
Q9. A ball of mass 0.5 kg is thrown vertically upward with speed 10 m/s. Using energy concepts, determine the maximum height reached. Explain how kinetic and potential energy change during the motion. (Take g = 9.8 m/s².)
Answer:
Initial kinetic energy: KE_initial = ½ m v² = 0.5 × 0.5 × 10² = 25 J.
At maximum height, the ball’s speed is zero, so KE_final = 0 and all initial kinetic energy has converted into gravitational potential energy (PE = m g h) (ignoring air resistance). Equate: m g h = 25 J. So h = 25 / (0.5 × 9.8) = 25 / 4.9 ≈ 5.10 m.
During the upward motion, kinetic energy decreases while potential energy increases by the same amount, keeping total mechanical energy constant (in absence of air resistance). This illustrates conversion: KE → PE. On the way down, the reverse happens: PE → KE, increasing the speed as height decreases. Energy conservation provides a simple way to find maximum height without solving kinematic equations.
Q10. A worker uses a machine to pull a heavy crate up a frictionless ramp of height 3 m. The crate’s mass is 50 kg. Another worker carries an identical crate straight up a staircase of the same vertical height 3 m. Compare the work done against gravity in both cases, and discuss the role of power if the first worker takes 30 s while the second takes 15 s to reach the top.
Answer:
Work against gravity only depends on vertical displacement: W = m g h. For both workers: m = 50 kg, h = 3 m, g = 9.8 m/s². So W = 50 × 9.8 × 3 = 1470 J. Both do the **sam...
