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  • A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the work done by the applied force in 10 s,

 Q2 (a)  A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the  work done by the applied force in 10 s,

             

Answers (1)

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Using Newton's law we can write :

                                                    a\ =\ \frac{F}{m}

                                                          =\ \frac{7}{2}\ =\ 3.5\ m/s^2

The frictional force is given by :

                                                     f\ =\ \mu mg

                                                          =\ 0.1\times 2\times 9.8\ =\ 1.96\ N

Its direction will be opposite of the direction of the motion. Thus acceleration produced will be negative.

                                                   a\ =\ \frac{-1.96}{2}\ =\ -0.98\ m/s^2

Thus the net acceleration is  =   3.5  -   0.98  =   2.52 m/s2.

The total distance travelled is given by :

                                                s\ =\ ut\ +\ \frac{1}{2}at^2

or                                                  =\ 0\ +\ \frac{1}{2}(2.52)10^2\ =\ 126\ m

Hence the work done by applied force is given by :

                                              W\ =\ F.s\ =\ 7\times 126\ =\ 882\ J 

Posted by

Devendra Khairwa

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