A factory worker pushes a 29.5-kg crate a distance of 4.4 m along a level floor

Question

A factory worker pushes a 29.5-kg crate a distance of 4.4 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.29. Suppose the worker pushes downward at and angle of 26° below horizontal.

(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?


How much work is done on the crate by gravity?


(e) What is the total work done on the crate?

Explanation / Answer

m = 29.5 Kg
d = 4.4 m
uk = 0.29
= 26o

(a)
Let the force applied be F.
F*cos() = uk * N
Where, N = m*g + F*sin()
F*cos() = uk * [m*g + F*sin()]
Substituting Values,
F * cos(26) = 0.29 * [29.5 * 9.8 + F * sin(26)]
F = 108.65 N

(b)
Work done, = Force * Distance
Work done, = F * cos(26) * d
Work done, = 108.65 * cos(26) * 4.4 J
Work done, = 309.3 J

(c)
As they are moving at constant velocity !!
Work done by Friction = - Work done by Force = - 309.3 J

(d)
As there is no motion in the vertical direction !!
Work done by Normal Force = Work done by Gravity = 0

(e)
Total Work done = 0

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