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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 change in kinetic energy of the body in 10 s

Q2 (d)  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 change in kinetic energy of the body in 10 s

       

Answers (1)

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It is given that initial velocity is zero. The final velocity can be calculated by the equation of motion :

                                                    v\ =\ u\ +\ at

or                                                       =\ 0\ +\ (2.52)10

or                                                       =\ 25.2\ m/s

Thus change is kinetic energy is :

                                                 \Delta K\ =\ \frac{1}{2}mv^2\ -\ \frac{1}{2}mu^2

or                                                           =\ \frac{1}{2}\times 2(25.2)^2\ -\ 0

or                                                            =\ 635\ J

Posted by

Devendra Khairwa

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