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

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To determine the total work done on a 25 kg crate pushed 5 meters at constant velocity, the required force is calculated using the coefficient of kinetic friction (0.23), resulting in a pushing force of approximately 57.5 N. The work done by the worker on the crate, calculated as force times distance, equals 287.5 J. The work done by friction is negative, amounting to -287.5 J, as it opposes the applied force. The normal force and gravitational force do not contribute to work done on the crate since they act perpendicular to the direction of motion. Therefore, the total work done on the crate is zero, as the positive work by the worker is balanced by the negative work done by friction.
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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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