As shown, a 58.0 kg crate is pulled up a theta = 50.0 degree incline by a pulley
ID: 1826852 • Letter: A
Question
As shown, a 58.0 kg crate is pulled up a theta = 50.0 degree incline by a pulley and motor system. Initially at rest, the crate is pulled s = 3.75 m up the incline. Undergoing constant acceleration, the crate reaches a speed of 2.60 m/s at the instant it has traveled this distance. Power supplied to the crate when friction is considered Considering the coefficient of kinetic friction, mu k = 0.13, determine the power that the motor must supply to the crate the instant the crate travels a distance of 3.75 m . Power supplied to the motor when efficiency is considered If the motor has an efficiency of epsilon = 0.88 , what power must be supplied to the motor to raise the crate?Explanation / Answer
v = at
t = v/a
s = 1/2 a * t* t = 0.5*a*(v/a)*(v/a)
= 0.5*v^2/a
=>a = 0.5*v^2/s = 338/375
Net forward force = ma = 58*a = 52.2773N
Friction force = u*Normal force = u*mg*cos50 = 0.13**50*9.8*cos50 = 47.4969N
Total backward force = friction + mgsin(50) = 47.4969+58*9.8*sin(50) = 482.9165
Total forward force = backward + net force = 482.9168+52.2773 = 535.1938N
Power supplied = force * velocity = 535.1938*2.60 = 1391.504W
Power supplied to motor = 1391.504/0.88 = 1581.2548W
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