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 30 ∘ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.26.
1.What magnitude of force must the worker apply to move the crate at constant velocity?
2.How much work is done on the crate by this force when the crate is pushed a distance of 5.0 m ?
3.How much work is done on the crate by friction during this displacement?
4.How much work is done by the normal force?
5.How much work is done by gravity?
6.What is the total work done on the crate?
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According to the concept of the work energy and power
Given that
Mass m=29.8 kg
Distance L =5 m
Angle =30°
Coefficient of friction =0.26
Now we find the magnitude of the force must the worker
Force F=mgsinα=29.8*9.8*sin30=146.02 N
Now we find the work done by the force
Work W1=146.02*5=730.1 J
Now we find the work is done on the create by friction
Wf=-(0.26*29.8*9.8*cos30°)=-65.8 J
Now we find the work done by the normal force
Wn=0
Now we find the work is done by gravity
Wg=29.8*9.8*5*sin(30)=730.1 J
Now we find the total work done on the create
Total work done W=Wf+Wg
=(730.1-65.8)=664.3 J
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