A 6-kg box is being lifted by means of a light rope that is threaded through a s

Question

A 6-kg box is being lifted by means of a light rope that is threaded through a single, light, frictionless pulley that is attached to the ceiling.

(a) If the box is being lifted at a constant speed of 2.0 m/s, what is the power delivered by the person pulling on the rope?

0.118 kW This solution was easy:

F = mg

F = 6*9.81

F = 58.86

P = F * v

P = 58.86 * 2

P = 117.72W

-OR-

P = 0.118 kW

(b) If the box is lifted, at constant acceleration, from rest on the floor to a height of 1.5 m above the floor in 0.42 s, what average power is delivered by the person pulling on the rope? kW

574 W

-OR-

0.574 kW

Please include the steps to solve part b much as I did in part a. How do I get from the question to the solution of .574kW?

Explanation / Answer

h = 1.5 m t = .42 s h = 1/2 * a * t^2====> a = 2h / t^2 = 3 / .1764 = 17.00 m/s^2 final velocity, v = sqrt(2 * a * h) = sqrt(2*17*1.5) = sqrt(51) kinetic energy of the box after lift = 1/2 mv^2 = .5 * 6 * 51 = 153 J potential energy gained by the box = m * g * h = 6 * 9.81 * 1.5 = 88.29 J total energy of box after lift = 153 + 88.29 = 241.29 J thus, the average power delvered by man is 241.29/0.42 = Energy delivered/time taken = 574.5 W = .5745 kW

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