How Is Work Calculated When Pushing a Desk Across a Rough Surface?

AI Thread Summary
Amadeus is calculating the work done while pushing a desk across a rough surface, considering both static and kinetic friction coefficients. The static friction coefficient is 0.800, while the kinetic friction coefficient is 0.620. After determining the force required to move the desk, he mistakenly uses the static friction coefficient for his calculations. The discussion suggests using the kinetic friction coefficient instead, as it applies once the desk is in motion. This adjustment is crucial for accurately calculating the work done during the push.
terichristine
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Homework Statement


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...)?


Homework Equations



u*mg
f=ma
w=d*f
D=1/2at^2

The Attempt at a Solution



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 that's wrong. I don't know what I am doing wrong. Thanks.
 
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terichristine said:

Homework Statement


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...)?


Homework Equations



u*mg
f=ma
w=d*f
D=1/2at^2

The Attempt at a Solution



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 that's wrong. I don't know what I am doing wrong. Thanks.

Try using mu k, coefficient of kinetic friction, because the only work done is while the force displacing the box is acting. It said this in the hint. So 0.62 instead of 0.80
 
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