Aworker pushes a 1.50 x 10^3 n crate with a horizontal force of 345 n a distance of 24.0 m. assume the coefficient of kinetic friction between the crate and the floor is 0.220.

a) how much work is done by the worker on the crate?
b) how much work is fone by the floor on the crate?
c) what is the net work done on the crate?

a) 8280 Joules

b) -7920 Joules

c) 360 Joules

Explanation:

Knowns:

Wt = 1500 [N] ==> Weight of crate

F = 345 [N] ==> Force applied on crate

μk = .220 ==> Coefficient of friction between crate and floor

Uknowns:

Wworker ==> Work done by worker on the crate

Wfloor ==> Work done by floor on the crate

Wnet ==> Net work done on the crate

First off, let´s explain what work is. Work is a physical quantity that results from the dot product of Force and displacement. Since it is a dot product or scalar product, the force and displacement must be parallel.

Where W = Work, F = Force, d = displacement

Now that we know what work is, let´s solve the problem:

a) Wworker = 345 [N] * 24 [m] = 8280 [J]

b) To solve b, we need to find out what is the Force of friction acting on the crate. We know the coefficient of friction and the weight of the crate. The equation for Force of friction is:

Where N is the normal force acting on the Crate. Since we know the weight of the crate, we know that the floor must be exerting the same force on the crate as the crate is exerting on the floor (Newtons third law). Therefore

Also, the Force of Friction is acting in opposite direction of the worker´s force, this means any work done by friction on the crate will be negative:

Wfloor = -μNd = .22*1500[N]*24[m] = -7920 [J]

c) Finally, the net work done is the sum of any work done on the system.

Wnet = Wworker + Wfloor

Wnet = 8280 [J] + (-7920[J])

Wnet = 360 [J]

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