A factory worker pushes a 30.2kg crate a distance of 5.0m along a level floor at

Question

A factory worker pushes a 30.2kg crate a distance of 5.0m 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.

A) What magnitude of force must the worker apply to move the crate at constant velocity?

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) sum forces in the y

N - m g - F sin theta = 0

N = 30.2*9.81 + F*sin(30 degrees)

sum fores in the x

Fcos(30) - friction = 0

Fcos(30) = friction

F*cos(30) = 0.26*( 30.2*9.81 + F*sin(30 degrees))

F=104.65 N

b) W = F d cos theta = 104.65*5*cos(30 degrees)=453 J

c) W friction = -friction d = -1*0.26*( 30.2*9.81 + 104.65*sin(30 degrees))*5=-453 J

d) angle for normal = 90

so W = 0

e) angle for gravity = 0 so W = 0

f) network = sum of works = 453-453 = 0

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