Question

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

Answer 1

Answer:

The workdone by P is $W_p=2.2J$

Explanation:

Generally workdone is mathematically represented as

              $W = F *d$

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

                $\Delta KE = KE_2 - KE_1$

Substituting 5.6 J for $KE_1$ and 4.0 J for $KE_2$

              $\Delta KE = 5.6 - 4$

                         $= 1.6J$

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

       The workdone by the frictional force is

                  $W_F = -F_f * d$

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

substituting 1.2 N for $F_f$  and 0.5 m for d

               $W_F = -1.2*0.5$

                     $=-0.6J$

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

            Workdone by Force P ($W_P$) = The total workdone($W_T$)  - $W_F$

Substitution values

                    $W_p = 1.6 -(-0.6)$

                          $W_p=2.2J$