What is the total work done on the crate with friction?

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SUMMARY

The discussion focuses on calculating the total work done on a 25Kg crate pushed 5m along a level floor with a coefficient of kinetic friction of 0.23. The worker must apply a force of 5.67N to overcome friction. The work done by this force is 28.35J, while the work done by friction is -28.35J, resulting in zero net work from friction. The normal force and gravity do not contribute to work since they act perpendicular to the direction of motion. The total work done on the crate is 0J.

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A worker pushes 25Kg crate a distance of 5m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic frication between the crate and the floor is 0.23.

(a) What magnitude of force must the worker apply?
(b) How much work is done on the crate by this force?
(c) How much work is done on the crate by frication?
(d) How much work is done on the crate by normal force?
(e) How much work is done on the crate by the gravity?
(f) What is the total work done on the crate?
 
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calculate the WEIGHT of the crate (downwards force).
Use the coeff of friction to calculate the force needed to push the crate...go from there
 

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