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PhysicsWork Power Energy
The power of a crane, which lifts a mass of 1000 kg to a height of 20 m in 10 s is : (g = 9·8 m/s²)
Options
1
19·6 W
2
39·2 W
3
19·6 kW
4
39·2 kW
Correct Answer
Option 3 : 19·6 kW
Step-by-Step Solution
1

Work Done against gravity:

\(W = mgh = 1000 \times 9.8 \times 20 = 196000 \text{ J}\)

2

Power:

\(P = \dfrac{W}{t} = \dfrac{196000}{10} = 19600 \text{ W} = \mathbf{19.6 \text{ kW}}\) ✓

Alt method — P = Fv:

F = mg = 9800 N, v = 20/10 = 2 m/s

P = 9800 × 2 = 19600 W = 19.6 kW ✓

Theory: Work, Power & Energy
1. Work Done

Work = Force × displacement × cosθ. For vertical lifting: W = mgh (θ = 0°). Work is scalar, measured in Joules. The work done against gravity depends only on height, not on the path taken — gravity is a conservative force.

\(W = mgh\) (lifting against gravity)

2. Power — Rate of Work

Power is how fast work is done. Same work in less time = more power. A 100W bulb uses 100J every second.

\(P = \dfrac{W}{t} = \dfrac{mgh}{t}\)

\(P = Fv\) (force × velocity)

📌 1 W = 1 J/s (SI unit)

📌 1 kW = 1000 W

📌 1 HP = 746 W

📌 1 kWh = 3.6 × 10⁶ J (electricity billing unit)

3. Work-Energy Theorem

Net work done = Change in KE: W_net = ΔKE. When lifting at constant speed, ΔKE = 0, so all work by crane goes to gravitational PE. Work done by crane = mgh = PE gained by load.

4. Power of Machines — Real Life

📌 Typical car engine: 50–150 kW

📌 This crane: 19.6 kW ≈ 26 HP

📌 Human cycling: ~200–400 W

📌 100W bulb: consumes 100J every second

📌 Electric kettle: ~2000 W = 2 kW

5. Pump Power Formula

For a pump lifting water of density ρ, volume V to height h in time t:

\(P = \dfrac{\rho V g h}{t} = \rho g h \times \text{flow rate}\)

This is frequently asked in NEET — same formula as the crane, with density × volume replacing mass.

Frequently Asked Questions
1. Why kW and not W?
W = 196000 J, t = 10 s → P = 19600 W = 19.6 kW. Option 19.6 W would only make sense for lifting ~1g to 20m in 10s. A real crane lifting 1000 kg absolutely requires kilowatts.
2. What is the speed of the lifted mass?
v = h/t = 20/10 = 2 m/s. At this speed, P = Fv = mg × v = 9800 × 2 = 19600 W = 19.6 kW — same answer as W/t method.
3. Does work done change if crane takes 20s instead of 10s?
No! Work done W = mgh = 196000 J — same regardless of time. But power halves: P = 196000/20 = 9800 W = 9.8 kW. More time → less power needed.
4. What is 1 kWh?
1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J. This is the unit your electricity meter measures. If this crane runs for 1 hour, it uses 19.6 kW × 1 h = 19.6 kWh of electrical energy.
5. How is power related to efficiency?
η = (useful output power)/(input power). If crane is 80% efficient, input power = 19.6/0.8 = 24.5 kW. Only 19.6 kW actually lifts the load; remaining 4.9 kW is wasted as heat in motors, ropes, etc.
6. What is the work done by gravity in this problem?
Work done BY gravity = −mgh = −196000 J (negative, since gravity acts downward but displacement is upward). Work done by crane = +196000 J. Net work = 0 → constant speed (ΔKE = 0) ✓.
7. Can power be negative?
Yes. When braking force acts opposite to motion, work done by that force is negative, so power = W/t is negative. This means energy is being absorbed rather than supplied (like regenerative braking in electric vehicles).
8. What is the dimension of power?
P = W/t = (Force × displacement)/time = (MLT⁻²)(L)/T = ML²T⁻³. So [P] = ML²T⁻³. Verify: 1 W = 1 kg·m²/s³ = 1 J/s.
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