Scientific Conception of Work — Long Answer Questions (Class 9 Physics)
Medium Level (Application & Explanation)
Q1. Define work in scientific terms and explain the two necessary conditions for work to be done. Give everyday examples to illustrate both conditions.
Answer:
In science, work is done when a force acts on an object and the object undergoes displacement in the direction of that force.
The two necessary conditions are: (1) a force must be applied, and (2) there must be displacement of the object in the direction of the force. If either is missing, no work is done.
Example 1: Pushing a pebble — you apply a force and the pebble moves, so work is done.
Example 2: Lifting a book — your upward force causes upward displacement, so work is done against gravity.
Example 3: Pushing a locked door — you apply force but there is no displacement, so no work is done on the door even though you feel tired.
These examples help connect the definition to daily life and emphasize the need for both force and displacement.
Q2. Explain why pushing a wall does not count as work done, even though a person exerts a force. Mention what actually happens to the energy the person uses.
Answer:
When you push a wall and it does not move, there is a force but no displacement, so by the scientific definition no work is done on the wall.
Physically, the wall’s point of contact does not shift, so the term W = F × s (where s is displacement) gives W = 0.
However, your muscles are active and consume chemical energy. That energy is converted into internal heat and some small internal work within muscles, making you feel tired.
Thus, energy is expended in your body, but no mechanical work is transferred to the wall.
This shows the difference between physiological effort and mechanical work in physics.
Q3. A student carries a heavy box horizontally at a constant height across the classroom. Is work being done on the box by the student? Explain clearly.
Answer:
If the student carries the box horizontally at a constant height, the force the student applies to support the box is vertical (upward) while the displacement is horizontal.
Because the displacement is perpendicular to the support force, the component of force in the direction of displacement is zero, so work done by that upward force on the box is zero.
If the student exerts any small horizontal push to overcome friction and the box moves, then work is done by that horizontal component.
Also, the student’s muscles use chemical energy and produce heat, so the person gets tired even though mechanical work on the box (by the vertical force) is zero.
In short, no work is done by the upward force on the box, but the body still expends energy internally.
Q4. Describe how gravitational force does work on a falling object. What happens to the object’s energy during the fall?
Answer:
When an object falls, gravity acts downwards and the object’s displacement is also downwards, so gravity does positive work on the object.
This work increases the object’s kinetic energy because gravitational potential energy is converted into kinetic energy as the object speeds up.
If an object of mass m falls through height h, its loss in potential energy (mgh) appears as a gain in kinetic energy (½mv²), ignoring air resistance.
Thus, work done by gravity is the mechanism for energy transfer from potential to kinetic form.
If air resistance is present, some energy also converts into thermal energy of air and object, but gravity still performs work that decreases potential energy and increases kinetic or other energy forms.
Q5. Explain how the angle between force and displacement affects the work done. Use the example of pulling a trolley with a rope inclined to the horizontal.
Answer:
When a force acts at an angle θ to the direction of displacement, only the component of the force along the displacement does work. The work is given by W = F × s × cosθ, where F is the force magnitude and s is displacement.
Example: pulling a trolley with a rope at angle θ above horizontal. The horizontal component F cosθ moves the trolley, so work done equals F cosθ × s.
If θ = 0° (force along displacement), cosθ = 1 and work is maximum (W = F × s). If θ = 90° (force perpendicular), cosθ = 0 and no work is done by that force.
This explains why pulling slightly upward reduces the horizontal force component and thus may reduce or change the work done against friction and movement.
High Complexity (Analytical & Scenario-Based)
Q6. A student pulls a suitcase with a force of 20 N using a rope inclined at 30° above the horizontal. The suitcase moves 5 m horizontally. Calculate the work done by the student and explain the reasoning.
Answer:
The work done depends on the component of the force along the displacement. The displacement is horizontal, so use the horizontal component of the force: F_horizontal = 20 N × cos(30°).
cos(30°) = 0.866, so F_horizontal ≈ 20 × 0.866 = 17.32 N.
Work done W = F_horizontal × displacement = 17.32 N × 5 m ≈ 86.6 J.
Interpretation: although the applied force is 20 N along the rope, only about 17.32 N acts in the direction of motion, so the mechanical work for displacement is 86.6 joules.
The vertical component does not contribute to horizontal work but may reduce the normal force and friction.
Q7. A worker holds a heavy beam motionless above the ground for two hours. From a physics perspective is work done on the beam? Discuss carefully and mention what happens to the worker’s energy.
Answer:
From a strict physics viewpoint, since the beam does not move, there is no displacement, and therefore no work is done on the beam by the worker’s supporting force (W = F × s = 0).
However, the worker’s muscles continuously contract to maintain the beam’s position, consuming chemical energy within the body. This energy is mostly turned into heat and some small internal work in muscles and tissues.
So, even though the beam receives no mechanical work, the worker expends metabolic energy and becomes tired.
This example highlights the difference between mechanical work on an object and the physiological energy used by a living body.
Q8. A person pushes a sled on level ground with force F but the sled does not move because static friction equals the push. Analyze which forces do work, whether energy is transferred, and what happens to the person’s energy.
Answer:
If the sled does not move, its displacement is zero, so no mechanical work is done on the sled by any external force (applied force, friction, normal, gravity) because W = F × s = 0 for each.
There is no transfer of mechanical energy to the sled. However, the person exerts muscular effort and uses chemical energy, which is converted into heat in the muscles and possibly some small internal work. The person will get tired.
The static friction force acts to balance the applied force, but with no displacement there is no frictional work either.
Thus, although forces act, no mechanical work is done on the sled; energy from the person is dissipated inside their body as heat.
Q9. A bird flies steadily at constant speed and altitude. Explain whether the bird’s wings perform work on the air and whether the bird does mechanical work to stay aloft.
Answer:
A bird’s wings push air downwards and backwards; this causes displacement of air in the direction of the applied force, so the wings do work on the air. That work transfers energy to the air (kinetic and some thermal energy).
To remain at constant altitude, the bird must produce lift equal to its weight. The wings do work to redirect airflow and maintain lift, and muscles supply chemical energy to flap the wings.
Even at constant speed and altitude, the bird does mechanical work to overcome air resistance and to keep airflow patterns that provide lift. Energy from muscles converts into mechanical work on air and then partly into heat.
So yes, the bird’s wings perform work on the air, and the bird expends energy continuously to stay aloft.
Q10. A car engine is running while the car is parked and the brakes hold it in place. Is mechanical work being done by the engine on the car? Explain what happens to the energy produced by the engine.
Answer:
If the car does not move, there is no displacement of the car’s center of mass, so the engine does no mechanical work on the car as a whole (W = F × s = 0 for net external force causing displacement).
Internally, the engine generates forces on pistons and moving parts, and chemical energy from fuel is converted into internal mechanical work, which is then mostly dissipated as heat through friction, exhaust, and the cooling system.
The brakes may also absorb small movements or vibrations as heat. Some work is done on gases in cylinders and exhaust, but this does not result in displacement of the car.
Hence, while the engine is active and energy is consumed, no useful external mechanical work moves the car; energy is lost as heat and sound.
Notes:
In all answers, emphasize the two essential conditions for work: force and displacement in the direction of force.
Remember the formula for work when force and displacement are in the same straight line: W = F × s; when angled, W = F × s × cosθ.
