The power required to lift a mass of 8·0 kg up a vertical distance of

The power required to lift a mass of 8·0 kg up a vertical distance of 4 m in 2 s is (taking acceleration due to gravity as 10 m/s²):

[amp_mcq option1=”80 W” option2=”160 W” option3=”320 W” option4=”640 W” correct=”option2″]

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UPSC NDA-1 – 2023

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The power required to lift an object is equal to the work done in lifting the object divided by the time taken. Work done in lifting is equal to the force required to lift it multiplied by the vertical distance lifted. The minimum force required to lift a mass vertically at a constant speed or to overcome gravity is equal to its weight.

– Weight (Force) = mass × acceleration due to gravity (W = mg).
– Work done (W) = Force × distance (W = Fd).
– Power (P) = Work done / Time (P = W/t).

Given: mass m = 8.0 kg, distance d = 4 m, time t = 2 s, acceleration due to gravity g = 10 m/s².
Force required to lift the mass = W = mg = 8.0 kg × 10 m/s² = 80 N.
Work done = Fd = 80 N × 4 m = 320 Joules.
Power = Work done / Time = 320 J / 2 s = 160 Watts.

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