Question

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity (zero acceleration) by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (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 friction? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?

Answer 1

Answer:

a) F = 73.5 N , b) W = 330.75 J , c) W = - 330.75 J , d)  W = 0 , e) W = 0 J

Explanation:

Let's write the equilibrium equations for this system

X axis

            F –fr = 0

            F = fr

Y Axis

            N - W = 0

            N = W

The equation for friction force is

            fr = μ N

            fr = μ m g

a)     F = fr = μ m g

       F = 0.25 30 9.8

       F = 73.5 N

b) the work is

         W = F x cos θ

The force is in the direction of displacement, so the angle is zero

           W = F x

           W = 73.5 4.5

           W = 330.75 J

c) the friction force has the same magnitude, but opposite direction, therefore the work is

              W = - 330.75 J

d) The normal force is perpendicular to the movement, so its angle is 90º

             Cos 90 = 0

             W = 0

e)

         W = W₁ + W₂

          W = 330.75 -330.75

          W = 0 J