Rate of Doing Work — Long Answer Questions (Class 9 Physics)
Medium Level (Application & Explanation)
Q1. Define power and derive the relation P = W/t. Explain the unit of power and why the watt is used.
Answer:
Power is the rate at which work is done or energy is transferred. If some work W is done in time t, then the faster the work is done, the more power is delivered.
Derivation: If an amount of work W is completed in time t, the average rate is work divided by time. So, P = W / t. This tells us how many joules of work are done each second.
Unit: Work is measured in joules (J) and time in seconds (s), so power has units J/s.
The SI unit J/s is named watt (W) after James Watt. We use watt because it is convenient and widely accepted; 1 W = 1 J/s.
For larger powers we use kilowatt (kW) where 1 kW = 1000 W. This is helpful for electrical appliances and engines.
Q2. Two children of equal weight climb the same rope to a height of 8 m. Child A takes 15 s and Child B takes 20 s. Who is more powerful? Show the reasoning.
Answer:
Both do the same work because work = weight × height and both have equal weight and climb the same height (8 m). Let that common work be W.
Power of A: P_A = W / 15 s. Power of B: P_B = W / 20 s.
Since 15 s < 20 s, P_A > P_B. This means A is more powerful because A did the same work in less time.
Example explanation: If W = 400 N × 8 m = 3200 J, then P_A = 3200 / 15 ≈ 213.3 W and P_B = 3200 / 20 = 160 W.
Conclusion: Higher power means doing the same work in less time.
Q3. A lamp consumes 1000 J of electrical energy in 10 s. Calculate its power. Explain how to convert this power into kilowatts and why kilowatts are useful for household appliances.
Answer:
Using P = W / t, with W = 1000 J and t = 10 s, P = 1000 / 10 = 100 W. So the lamp’s power is 100 watts.
To convert to kilowatts, divide by 1000. So 100 W = 0.1 kW.
Kilowatts are useful because many household devices (electric heaters, refrigerators, motors) have powers in the hundreds or thousands of watts; using kW gives simpler numbers (for example, 1.5 kW instead of 1500 W).
Electricity bills often use kilowatt-hour (kWh) as an energy unit, which is power (kW) multiplied by time (hours). Knowing appliance power in kW helps estimate energy consumption and cost.
In summary, the lamp’s power is 100 W (0.1 kW) and kilowatts simplify practical energy calculations.
Q4. A boy weighing 50 kg runs up 45 steps in 9 s. Each step is 15 cm high. Calculate his power and explain each step clearly.
Answer:
First find the boy’s weight (force due to gravity): mass m = 50 kg, take g ≈ 10 m/s², so weight = mg = 50 × 10 = 500 N (this is the force he lifts against gravity).
Total height climbed: 45 steps × 15 cm per step = 45 × 0.15 m = 6.75 m.
Work done = weight × height = 500 N × 6.75 m = 3375 J. This is the energy needed to raise his body.
Power = work / time = 3375 J / 9 s = 375 W. So the boy’s power output is 375 watts.
This calculation assumes all work goes into gaining gravitational potential energy and ignores losses like heat or additional muscular work, so actual energy used by the body will be higher.
Q5. What is average power? Give a situation where power varies with time and explain how average power is useful there.
Answer:
Average power is the total work done or total energy transferred divided by the total time taken: Average Power = Total Energy / Total Time.
Example situation: a worker uses a hand drill that runs slowly at first, then speeds up, then slows again. The instantaneous power changes at each moment because work rate is not constant.
If during a 10-minute job the drill uses 12,000 J of energy in total, average power = 12,000 J / 600 s = 20 W. This single number represents the overall rate of energy use.
Average power is useful for planning and comparisons: it helps estimate energy consumption, battery life, or heating effects when the device’s power is not steady.
It smooths out fluctuations to give a meaningful overall measure of performance.
High Complexity (Analytical & Scenario-Based)
Q6. Two machines A and B do the same amount of useful work. Machine A completes the job in 10 s but wastes a lot of heat; machine B takes 20 s but wastes less heat. Compare their power outputs and efficiencies, and explain which machine is preferable in different situations.
Answer:
Both machines do the same useful work W. Machine A’s useful power output is P_A = W / 10 s = 0.1 W per second of work unit and machine B’s useful power output is P_B = W / 20 s which is half of A’s useful power. So A has higher useful power.
Efficiency = useful work output / total energy input. If A wastes more heat, its efficiency is lower than B’s. For example, if A’s input energy is 2W for every 1W useful, efficiency = 50%; if B’s input is 1.25W for 1W useful, efficiency = 80%.
Preference depends on needs:
If speed matters (urgent tasks), A is preferable because higher power finishes job faster despite lower efficiency.
If energy saving or lower running cost matters, B is preferable because higher efficiency means less wasted energy and lower heat.
Real installations balance power and efficiency, often using powerful machines for short bursts and efficient machines for continuous operation.
Q7. A lift raises a 600 kg load through 4 m in 5 s. Calculate: (a) the mechanical power required (output), (b) the electrical input power if the motor efficiency is 80%, and (c) the current drawn from a 230 V supply. Show calculations and explain assumptions.
Answer:
(a) Calculate weight: mg = 600 kg × 10 m/s² = 6000 N. Work = force × height = 6000 N × 4 m = 24,000 J. Mechanical power output = 24,000 J / 5 s = 4,800 W.
(b) Motor efficiency η = 80% = 0.8. Electrical input power = mechanical output / η = 4,800 W / 0.8 = 6,000 W. This accounts for energy losses (heat, friction).
(c) Current from a 230 V supply: I = Power / Voltage = 6,000 W / 230 V ≈ 26.09 A.
Assumptions: g = 10 m/s², no extra frictional losses beyond the efficiency already given, steady lifting speed. These figures show the motor must deliver substantial power and draw a high current.
Q8. Explain why using a high-power appliance on weak wiring is unsafe. Use a 2 kW heater connected to a 230 V supply as an example and discuss fuses and overheating.
Answer:
A 2 kW heater at 230 V draws current I = P / V = 2000 / 230 ≈ 8.7 A. If the wiring or socket is rated for less current (for example 5 A), the wire will carry more current than it safely can.
Excess current causes the wire to heat due to I²R losses (power dissipated as heat in wire resistance). Overheating can melt insulation and lead to short circuits or fire.
A fuse or circuit breaker protects the circuit by interrupting current if it exceeds safe limits. If a circuit is overloaded by multiple high-power appliances, the fuse should blow or breaker trip, preventing damage.
Using appliances within the wiring’s rated capacity and ensuring proper earthing prevents hazards. For high-power devices, use dedicated circuits and appropriate wire gauge and protective devices.
Q9. Two students A and B climb to the same height of 5 m. A has mass 40 kg and takes 10 s; B has mass 50 kg and takes 12 s. Who has greater power? Show calculations and explain why a heavier person taking longer can still have more power.
Answer:
Convert masses to weights (mg): A: 40 kg × 10 m/s² = 400 N. B: 50 kg × 10 m/s² = 500 N.
Work done by A = 400 N × 5 m = 2000 J. Work done by B = 500 N × 5 m = 2500 J.
Power of A = 2000 J / 10 s = 200 W. Power of B = 2500 J / 12 s ≈ 208.33 W.
Even though B took longer, B did more work because of greater weight. Therefore B’s power (≈208.3 W) is slightly higher than A’s (200 W).
This shows power depends on both work and time; a heavier person can have greater power if the extra work they do compensates for taking more time.
Q10. Design a simple experiment to measure the power output of a student climbing a staircase. Describe the procedure, the calculations, and possible sources of error with improvements.
Answer:
Procedure: Measure the vertical height (h) of the staircase (use tape measure). Ask the student’s mass (m). Time the climb from bottom to top using a stopwatch to get t (repeat several times and take average). Ensure student climbs at a steady pace and only carries negligible extra weight.
Calculations: Weight = mg (use g = 10 m/s²). Work done = mg × h (this raises the student’s potential energy). Power = work / t = mg h / t. Report power in watts.
Sources of error: ignoring extra energy spent due to horizontal motion, bending knees, or inefficiencies; using g = 9.8 vs 10; inaccurate height or stopwatch timing; variations in climb speed.
Improvements: use a more accurate stopwatch or photogate for timing, measure height precisely, average more trials, and account for metabolic inefficiency if comparing to caloric energy. Clearly state assumptions so results are interpreted correctly.
