A factory worker pushes a 28.5 kg crate a distance of 5.0 m along a level floor

Question

A factory worker pushes a 28.5 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 0.24.

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

Given that

A crate of mass (m) =28.5kg

Distance moved along the floor (d) =5m

The angle of elevation with the horizontal (theta) =30degrees

The coefficient of kinetic friction between the crate and floor is (uk) =0.24.

Considering the free body diagram we can consider vertical direction as

FN =Fsin30+mg

Then the frictional force is fk =ukFN =uk(Fsin30+mg)

Now along the horizontal direction is

Fcos30=fk

Fcos30 =uk(Fsin30+mg)

Fcos30 =(0.24)(Fsin30+28.5kg*9.81)

0.866F=0.12F+67.100

0.746F =67.100

         F =67.100/0.746 =89.946N

b)

The work done by the crate is W =F*distance =89.946N*5m =449.731J

c)

Work done by the friction is

Wf =fk.d =uk(Fsin30+mg)*dcos180 =(0.24)(89.946Nsin30+28.5*9.81)(5)cos180 =-449.731J

d)

The work done by the gravity is

Wg =mgdcos90 =(28.5)(9.81)(5)cos90 =0

e)

The total work done by the crate Wtotal =449.731-449.731+0 =0

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