A factory worker pushes a 28.9 kg crate a distance of 4.0 m along a level floor
ID: 1419502 • Letter: A
Question
A factory worker pushes a 28.9 kg crate a distance of 4.0 m along a level floor at constant velocity by pushing downward at an angle of 28 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 4.0 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
Since the velocity is constant, acceleration is zero.
Fcos(A) - uN = 0
Fsin(A) + mg - N = 0
Substitute for N
Fcos(A) - u(Fsin(A) + mg) = 0
Fcos(A) - usin(A) = umg
F = umg/(cos(A) - usin(A))
F = 0.26*28.9*9.8/(cos(28) - 0.26*sin(28))
F = 96.77 N >>>>>>>>>>>(a)
W = 96.77*cos(28)*4
W =341.77joules >>>>>>>>(b)
Work done by friction is also 341.77 joules.>>>>>>>>>(c)
How much work is done by the normal force?
Wnf= N*0 J 0 , 0 >>>>>>>>>>>>>>>>>>(d)
Normal forces perform no work, gravity perform no work since their directions are 90 degrees to the velocity.
How much work is done by gravity? >>>>>>>(e)
Wg=No displacement in the direction of Wg and hence the work done is zero.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Constant velocity means total forces = 0.
Total work done on the crate = total force x distance = 0 x 4 m = 0 joule>>>>>>>>>>>(f)
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