Power Output of a Motor
The crate has a mass of 150 kg and rests on a surface for which the coefficients of static and kinetic friction are us= 0.3 and uk = 0.2 respectively. If the motor supplies a cable force of F = (8t^2 + 20) N, where t is in seconds, determine the power output developed by the motor when t= 5s.
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The way the pulley system is organized, if F is the force exerted by the motor cable, the top cable attached to the crate has a force of 2F pulling it to the left and the bottom cable attached to the crate pulls it with a force of F to the left.
Hence, total force on the crate = 3F
F = (8t^2+20) N
The box starts moving to the left only when 3F equals to the opposing force due to static friction.
Static frictional force = m*g*Us = 150*9.8*0.3 = 441 N
Find the time at which the crate starts moving.
3F = 441
3(8t^2+20) = 441 ==> t = 3.9843 sec. = 4 sec.
The crate starts moving only at t =4 sec and then the accelerating force is:
Fa =
= 3F - Kinetic Frictional force
= 3*(8t^2+20) - 150*9.8*0.2 = [3*(8t^2+20) - 294] N
at t = 5 sec:
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3F = 660 N
Accelerating force= Fa = 366 N
Hence acceleration a = 366/150 = 2.44 m/s^2
Velocity = 2.44*1 = 2.44 m/s
Distance travelled s = (1/2)at^2 = 1.22 m
Work done by the motor in moving the crate by 1.22 m = 3F*s = 660*1.22 = 805.2 Joules.
Work done by the frictional force = 294*1.22 = 358.7 Joules
Total work done by the motor = 805.2 + 358.7 = 1164 Joules
This is the amount of power of power the motor has to develop per second
= 1164 Jouls/se = 1164 Watts = 1.164 kW (some numerical round-off from your 1.12 kW)