Question

a 15 kg block slides on a horizontal surface with an initial speed of 12 m/s. the speed of 5.0m/s
is attained after travelling a distance of 25m.(a) what is the change in kinetic energy on the box (b) how much work was done by friction on the block ?

Answer 1

a) The initial kinetic energy is:

$$K_i = \dfrac{mv_i^2}{2}$$

where $m=15kg$ and $v_i = 12m/s$ is the initial speed of the block. Similarly, the final kinetic energy is:

$$K_f = \dfrac{mv_f^2}{2}$$

where $v_f = 5m/s$ is the final speed of the block.

The change is:

$$\Delta K = K_f-K_i= \dfrac{mv_f^2}{2}- \dfrac{mv_i^2}{2} = \dfrac{m}{2}(v_f^2 - v_i^2)=\\ =\dfrac{15}{2}(5^2 - 12^2) = -892.5\space J$$

b) According to the energy conservation law, the work done by friction is equal to the negative change of the kinetic energy:

$$A = -\Delta K = 892.5\space J$$

Answer. a) -892.5 J, b) 892.5 J.