Amadeus pushes horizontally on a desk that rests on a rough wooden floor. The co

Question

Amadeus pushes horizontally on a desk that rests on a rough wooden floor. The coefficient of static friction between the desk and the floor is 0.800 and the coefficient of kinetic friction is 0.620. The desk has a mass of 94.0 kg. He pushes just hard enough to barely get the desk moving and continues pushing with that force for 6.00 s. What work does he do on the desk during that time (HINT: Since once movement starts, friction changes to sliding friction, which is less than maximum static friction, the desk will be accelerating. SO, use the 2nd law to find the acceleration, then find out far it went in 6 s....)?

I used:
u*mg=.8*94*9.8=736.96N
f=ma=736.96/94=acceleration=7.84
and displacement=1/2at^2=(.5)*(7.84)*(6^2)=141.2
w=d*f=141.2*736.96=104059 and thats wrong. I don't know what I am doing wrong. Thanks.

Explanation / Answer

use s, k to represent static and kinetic friction coefficients respectively

F = s mg

F - kmg = ma

a = (s - k)g

d = (1/2)at2

W = F*d = s mg * (1/2)(s-k)g) t2

= (1/2)ms(s - k)(gt)2

= 0.5*94*(0.8)(0.8-0.62)(9.8*6)2 = 2.34*104 Nm

your calculation for a is not correct.

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