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 29 below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.26. a)What magnitude of force must the worker apply to move the crate at constant velocity? Express your answer using two significant figures. b)How much work is done on the crate by this force when the crate is pushed a distance of 5.0 m ? Express your answer using two significant figures. c)How much work is done on the crate by friction during this displacement? Express your answer using two significant figures. 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)

Equating forces in vertical:

F*sin(29) + mg = N

Equating forces along horizontal:

F*cos(29) - u*N = 0

So, Fcos(29 deg) - 0.26*(F*sin(29 deg) + 29.8*9.8) = 0

So, F = 101.4 N <-------answer

b)

Work done = F*s = 101.4*5 = 507 J

c)

Work done by friction = -Fs = -507 J

d)

Work done by normal force = 0 <------ as no displacement along normal

e)

Work done by gravity = 0 <------ no displacement along vertical

f)

Total work done = 0 <---- net force is 0 . so total work is 0

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