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 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?

Explanation / Answer

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=mgsin=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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