WORK DONE BY A CONSTANT FORCE — Long Answer Questions
Medium Level (Application & Explanation)
Q1. Define work done by a constant force. Give the formula, unit, and a simple example to illustrate the concept.
Answer:
Work is said to be done when a force causes a displacement of an object in the direction of the force. For a constant force acting along the direction of displacement, the formula is W = F × s,
where W is work, F is force in newtons (N), and s is displacement in metres (m).
The SI unit of work is Newton metre (N·m), also called Joule (J); so 1 J = 1 N·m.
Example: If a force of 5 N moves an object 2 m in the same direction, then W = 5 N × 2 m = 10 J.
Remember: work requires both non-zero force and non-zero displacement in the same or component direction.
Q2. Explain positive, negative, and zero work with one example each and the reason behind each type.
Answer:
Positive work occurs when the force and displacement are in the same direction. Example: When you pull a trolley forward, the pulling force and trolley displacement are both forward → positive work is done.
Negative work happens when the force is opposite to the displacement. Example: When brakes apply a retarding force and the car moves forward while slowing down, the braking force is opposite to displacement → negative work, which removes energy from the car.
Zero work arises when either displacement is zero or force is perpendicular to displacement. Example: Pushing a wall hard with no movement → force exists but s = 0, so W = 0.
These types are decided by the relative direction of force and displacement.
Q3. A porter lifts a luggage of mass 15 kg from the ground to a height of 1.5 m. Taking g = 10 m/s², calculate the work done by the porter. Also state the work done by gravity.
Answer:
First find the lifting force (weight) the porter must overcome: F = mg = 15 kg × 10 m/s² = 150 N.
Displacement in the direction of this force is s = 1.5 m upward. Using W = F × s, work done by the porter is W_porter = 150 N × 1.5 m = 225 J. This is positive work because force and displacement are same direction (upward).
Work done by gravity during this upward displacement is equal in magnitude but negative: W_gravity = −225 J, since gravity acts downward while displacement is upward.
Net work done on the luggage (ignoring other forces) is zero if the porter lifts at constant speed (no change in kinetic energy).
Q4. Why does pushing a heavy immovable wall do no work, even though you apply a large force? Explain in simple terms with the formula.
Answer:
Work depends on both force and displacement, given by W = F × s. If you push a wall hard but the wall does not move, then displacement s = 0.
Substituting in the formula gives W = F × 0 = 0 J. So, despite a large force, no work is done because there is no displacement of the point where force acts.
Physically, energy you expend may be converted into internal energy in your muscles (you feel tired), but mechanical work on the wall is zero.
Key idea: Force alone is not enough; there must be displacement for mechanical work.
Q5. A force of 20 N moves a crate for 0.5 km along the floor in the direction of the force. Calculate the work done. Show unit conversion and final answer in joules.
Answer:
Convert displacement into metres: 0.5 km = 500 m.
Use W = F × s with F = 20 N and s = 500 m.
So W = 20 N × 500 m = 10,000 N·m = 10,000 J.
Therefore, the work done by the force is 10,000 J.
Note: Always convert units to SI (metres for displacement) before using the formula so the result is in Joules.
High Complexity (Analytical & Scenario-Based)
Q6. A porter carries a suitcase of mass 10 kg horizontally across a room 8 m long at constant height. Gravity acts downward and his upward force equals the weight. Calculate the work done by the porter and by gravity during this horizontal displacement. Explain why.
Answer:
The porter applies an upward force equal to the weight: F_up = mg = 10 kg × 10 m/s² = 100 N. However, the displacement is horizontal (8 m), not vertical.
Work done by a force depends on the component of force along the direction of displacement. Since the upward force is perpendicular to the horizontal displacement, its component along motion is zero. Thus W_porter = F_up × s × cos 90° = 0 J.
Gravity also acts downward (perpendicular to the horizontal displacement), so W_gravity = 0 J as well.
Conclusion: Carrying something horizontally at the same height involves no mechanical work against gravity, although the porter may get tired due to biological energy use; mechanically, work by gravity and the upward supporting force is zero.
Q7. A box is pulled to the right by a constant force of 20 N while friction of 8 N acts to the left. The box moves 6 m to the right. Calculate (a) work done by the pulling force, (b) work done by friction, and (c) net work done on the box. Explain signs and meanings.
Answer:
Work by pulling force (rightward): W_pull = F × s = 20 N × 6 m = 120 J (positive because force and displacement same direction).
Work by friction (leftward, opposite displacement): W_fric = (−8 N) × 6 m = −48 J (negative because friction opposes motion).
Net work is the algebraic sum: W_net = 120 J + (−48 J) = 72 J.
Interpretation: 72 J is the total mechanical energy supplied to the box after overcoming friction; it may increase the box’s kinetic energy or be stored elsewhere. Signs show energy added (positive) or removed (negative) by each force.
Q8. A crate is pulled along a rough floor by a constant force of 15 N at an angle of 30° above the horizontal for a distance of 4 m. Using components, calculate the work done by the pulling force. (Hint: only the horizontal component of force does work along horizontal displacement.)
Answer:
Resolve the pulling force into horizontal component: F_x = F cos θ = 15 N × cos 30° = 15 × (√3/2) ≈ 12.99 N.
Vertical component does no work for purely horizontal displacement (if vertical displacement is zero).
Work done by the pulling force along horizontal displacement: W = F_x × s = 12.99 N × 4 m ≈ 51.96 J.
Thus the pulling force does about 52 J of work on the crate in the direction of motion.
Key idea: When force is at an angle, only the component along displacement contributes to mechanical work.
Q9. A man lifts a crate 2 m up slowly, then lowers it back 2 m to the original point. Describe the work done by the man and by gravity during the lift and during lowering. What is the net work done over the complete trip? Explain why.
Answer:
During the upward lift (2 m), the man does positive work equal to W_up = mg × 2 m (he applies upward force). Gravity does negative work of −mg × 2 m.
During the lowering (2 m down), if he lowers slowly, his applied force is downward but smaller in sign; the man does negative work equal to −mg × 2 m (force is upward less than weight or controlled downward force), while gravity does positive work of +mg × 2 m.
Over the complete round trip, the man’s positive and negative works cancel: W_man_total = mg×2 + (−mg×2) = 0 J. Similarly W_gravity_total = 0 J.
Net work over complete trip is zero, because the crate returns to its initial position and there is no net change in its mechanical energy.
Q10. Design a simple classroom experiment to measure the work done by a constant horizontal force using a spring balance and a toy cart. Describe steps, measurements, calculations, and possible errors.
Answer:
Materials: toy cart, spring balance, measuring tape, stopwatch (optional), smooth horizontal surface.
Procedure: Attach spring balance to the cart. Pull the cart horizontally with a constant reading (say F) on the balance. Measure the horizontal displacement s using the tape (for example s = 3 m). Keep the force steady and ensure motion is along a straight line.
Calculation: Use W = F × s. If F = 10 N and s = 3 m, W = 30 J. Record units as Joules.
Observations: Repeat trials for different forces or distances and compare results.
Sources of error: varying force, frictional forces not accounted for, inaccurate reading of spring balance, non-horizontal motion, or inaccurate distance measurement. Mention these and suggest minimizing them by pulling smoothly, using a low-friction track, and averaging multiple trials.
