Power, Work and Friction Problem

AI Thread Summary
To calculate the power needed to push a 90 kg chest at 0.52 m/s on a horizontal surface with a coefficient of friction of 0.79, the required power is determined to be 360 Watts. The work done in pushing the chest over a distance of 9.0 m can be found using the formula Work = Force x Distance, where the force accounts for friction. The net force is zero when the velocity is constant, indicating that the applied force equals the frictional force. The discussion emphasizes the importance of understanding the balance of forces, particularly in scenarios involving friction. Overall, the calculations hinge on accurately determining the frictional force to solve for power and work.
K.Lewis
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Homework Statement


1, How much power is needed to push a 90 kg chest at 0.52 m/s along a horizontal floor where the coefficient of friction is 0.79?
2, How much work is done in pushing the chest 9.0m?


Homework Equations


Unsure sorry.


The Attempt at a Solution


I'm lost on how to solve either of these; but I do know that the answer to question 1 is 360 Watts. Don't know how it was found though...
 
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P= W/change in time

Since work= Fd, P=Fd/change in t= FVav

So then plug everything in for Power= Force x Average Velocity

For part 2, the normal force is just 90 is just perpendicular to the surface so plug in the numbers into Work = Force x Distance once you solve for distance.
 
Awesome, thank you!
But one more question.. How do I find that first force if force=mass x acceleration, but the velocity is constant for the object?
 
The Net force is zero when the velocity is constant, therefore there would be no acceleration. The only forces acting on the crate is the Normal force and gravity which cancel, and Force of friction, so just use that as your force. When you push it with contact, there is a force applied, but once you let go the only unbalanced force is friction. Since there is friction, the box would eventually come to a stop therefore the velocity wouldn't be constant as it slides, or it will slide on forever.
 
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