A factory worker pushes a 30.2kg crate a distance of 4.5m along a level floor at

Question

A factory worker pushes a 30.2kg crate a distance of 4.5m along a level floor at constant velocity by pushing downward at an angle of 29 below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25.

a-What magnitude of force must the worker apply to move the crate at constant velocity?

b-How much work is done on the crate by this force when the crate is pushed a distance of 4.5 m?

c-How much work is done on the crate by friction during this displacement?

d-How much work is done by the normal force?

e-How much work is done by gravity?

f-What is the total work done on the crate?

Explanation / Answer

(a) Applying the Newton's 2nd law of motion,

Fcas -(mg+Fsin) = ma ----------- (1)

since at constant speed, acceleration a = 0

Fcas = (mg+ Fsin)

F( 0.8746 - 0.1212 ) = 0.25*30.2*9.8

F = 98.22 N

the magnitude of force must the worker apply to move the crate at constant velocity is 98.22 N.

(b) the workdone W = F.S = 98.22* 4.5 = 441.99 J

(c) the frictional force, F = 0.25(30.2*9.8+98.22sin29)

F = 85.89 N

the workdone = 85.89*4.5 = 386.52 J

(d) the work is done by the normal force is zero.

(e) the work is done by gravity is zero.

(f)
the total work done W = 441.99 - 386.52 = 55.46 J

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