A factory worker pushes a 29.8 kg crate a distance of 5.0 m along a level floor

Question

A factory worker pushes a 29.8 kg crate a distance of 5.0 m along a level floor at constant velocity by pushing downward at an angle of 30 below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.26.

1.What magnitude of force must the worker apply to move the crate at constant velocity?

2.How much work is done on the crate by this force when the crate is pushed a distance of 5.0 m ?

3.How much work is done on the crate by friction during this displacement?

4.How much work is done by the normal force?

5.How much work is done by gravity?

Explanation / Answer

Consider the horizontal as x-axis and vertical as y-axis

Net Force along x-axis

Fnet,x = F*cos(theta) - kinetic frictional Force =0

F*cos(theta) = mu_k*Normal Force

F*cos(theta) = mu_k*N .....(1)

along y-axis
Fnet,y =0

N - (F*sin(theta)) - (m*g) = 0

N = F*sin(theta) + (m*g).....(2)

substitute (2) in (1) we get

then

F*cos(30) = 0.26*(F*sin(30)+(29.8*9.8))

Force is F = 103.16 N

2) Work done is W1 = F*cos(theta)*S = 103.16*cos(30)*5 = 446.7J

3)work done by friction is Wf = -446.7 J

4) WOrk done by normal Force is WN = N*S*cos(90) = 0 J

5) Work done by gravity is Wg = m*g*S*cos(90) = 0 J

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