A factory worker pushes a crate of mass 29.6kg a distance of 4.25m along a level

Question

A factory worker pushes a crate of mass 29.6kg a distance of 4.25m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.245.

1 What magnitude of force must the worker apply?
2 How much work is done on the crate by this force?
3 How much work is done on the crate by friction?
4 How much work is done by the normal force?
5 How much work is done by gravity?
6 What is the total work done on the crate?

Please make a step by step explanation for each questions ... with the literal expression (so I can make note out of this exercise and write a methodology )

thank you in advance

Explanation / Answer

a) What magnitude of force must the worker apply to move the crate at constant velocity? Total normal force = (mg ) Friction = ? (mg ) constant velocity implies the applied force F is equal to the frictional force. ? (mg ) = F 0.245. (29.6*9.8 ) = F F= 71.1 N b) How much work is done on the crate by this force when the crate is pushed a distance of 4.25 m? W= 71.1* 4.25 = 302.175 J c) How much work is done on the crate by friction during this displacement? Since 302.175 J is the work done against friction Wf= 302.175 J d) How much work is done by the normal force? Wnf= N*0 J e) How much work is done by gravity? Wg=No displacement in the direction of Wg and hence the work done is zero. f) What is the total work done on the crate? Wnet = 302.175 J

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