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Difficulty: EasyWork, Energy and Power

An electric crane lifts a load of 250 kg250\text{ kg} vertically upwards through a height of 12 m12\text{ m} in 10 s10\text{ s} at a constant speed. Taking the acceleration due to gravity g=10 m s−2g = 10\text{ m s}^{-2}, what is the useful output power of the crane in watts?

Answer: 3000 W

Answer

The useful output power of the crane is 3000 W3000\text{ W}.
The work done in lifting the load vertically is equal to the gravitational potential energy gained, W=mgh=250×10×12=30,000 JW = mgh = 250 \times 10 \times 12 = 30,000\text{ J}. Power is the rate of doing work, so P=Wt=30,00010=3000 WP = \frac{W}{t} = \frac{30,000}{10} = 3000\text{ W}.

Step-by-Step Solution

1
Identify the given values
Mass m=250 kgm = 250\text{ kg}, height h=12 mh = 12\text{ m}, time t=10 st = 10\text{ s}, acceleration due to gravity g=10 m s−2g = 10\text{ m s}^{-2}.
Extract values needed for work and power calculations.
2
Calculate the work done in lifting the load
W=mgh=250 kg×10 m s−2×12 m=30,000 JW = mgh = 250 \text{ kg} \times 10 \text{ m s}^{-2} \times 12 \text{ m} = 30,000\text{ J}.
The work done against gravity equals the gain in gravitational potential energy.
3
Calculate the power output
P=Wt=30,000 J10 s=3000 WP = \frac{W}{t} = \frac{30,000\text{ J}}{10\text{ s}} = 3000\text{ W}.
Power is defined as the rate at which work is done (P=WtP = \frac{W}{t}).

Key Concept

Power as the rate of doing work against gravity
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Topics

Work, Energy and Power

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