A block is pushed across a rough horizontal surface from point A to point B by a force P. The magnitude of the force of friction acting on the block between A and B is 1.2 N and points A and B are 0.5 m apart. If the kinetic energies of the block at A and B are 4.0 J and 5.6 J, respectively, how much work is done on the block by the force P between A and B

Respuesta :

Answer:

The workdone by P is [tex]W_p=2.2J[/tex]

Explanation:

Generally workdone is mathematically represented as

              [tex]W = F *d[/tex]

Where F is the force and d is the distance

  The total workdone = (Workdone by Force P + Workdone by frictional force )

The difference in kinetic energy

                [tex]\Delta KE = KE_2 - KE_1[/tex]

Substituting 5.6 J for [tex]KE_1[/tex] and 4.0 J for [tex]KE_2[/tex]

              [tex]\Delta KE = 5.6 - 4[/tex]

                         [tex]= 1.6J[/tex]

This change in kinetic energy of the block = The total kinetic energy

       The workdone by the frictional force is

                  [tex]W_F = -F_f * d[/tex]

the negative sign shown that the force is moving in the opposite direction

substituting 1.2 N for [tex]F_f[/tex]  and 0.5 m for d

               [tex]W_F = -1.2*0.5[/tex]

                     [tex]=-0.6J[/tex]

Making Workdone by Force P the subject in the above equation we

            Workdone by Force P ([tex]W_P[/tex]) = The total workdone([tex]W_T[/tex])  - [tex]W_F[/tex]

Substitution values

                    [tex]W_p = 1.6 -(-0.6)[/tex]

                          [tex]W_p=2.2J[/tex]

                   

                           

     

                 

The work done on the block by the force P between A and B is; 2.2 J

In this question, there will be work done by friction and work done by the applied force.

We are told that force of friction acting on the block between A and B is 1.2 N and points A and B are 0.5 m apart. Thus;

Frictional Force; F_f = 1.2 N.

∴ Work done by friction; W_f = -F_f × d

W_f = -1.2 × 0.5

W_f = -0.6 J (negative sign because frictional force is force that is opposing motion)

Now, we are told that the kinetic energies of the block at A and B are 4.0 J and 5.6 J respectively.

Since the block was pushed from point A to point B, then we can say that;

Change in kinetic energy = total work done = 5.6 - 4

Total work done = 1.6 J

I earlier said that we have work done by friction and work done by the applied force acting on this system. Thus;

Total work done = Work done by friction + work done by force

Plugging in the relevant values gives us;

1.6 = -0.6 + work done by force

work done by force = 1.6 + 0.6

work done by force = 2.2 J

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