A winch lifts 420 kg through 18 m in 24 s, from rest to 3.0 m/s, losing 1.6 × 10⁴ J. Take g = 9.81 m/s². (a) Find GPE and KE gains. (b) Find electrical input and average power. (c) At final speed input power is 18.0 kW and resistance 900 N. Find acceleration. (d) Explain average versus instantaneous power.

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GPE 7.42 × 10⁴ J; KE 1.89 × 10³ J; input 9.21 × 10⁴ J; average 3.84 kW; final acceleration about 2.33 m/s² upward. GPE gain = mgh = 420 × 9.81 × 18 = 7.42 × 10⁴ J. KE gain = ½mv² = ½ × 420 × 3.0² = 1.89 × 10³ J. Input energy supplies both gains plus losses: 74164 + 1890 + 16000 = 9.21 × 10⁴ J. Average power = 92054/24 = 3.84 kW. At final speed, motor force follows P = Fv: F = 18000/3.0 = 6000 N upward. Net force = 6000 − mg − 900 = 6000 − 4120 − 900 = 980 N upward, so acceleration = 980/420 = 2.33 m/s². Instantaneous power depends on force and speed at one instant; average power divides total energy by whole time, including low-speed starting and varying losses.

Winch energy and instantaneous power

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