Answer to Question #141113 in Physics for Hatdogi

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 change in kinetic energy on the box is easy to find: subtract final kinetic energy from initial:

$$\Delta KE=\frac{mv_f^2}{2}-\frac{mv_i^2}{2},\\\space\\ \Delta KE=\frac{15\cdot5^2}{2}-\frac{15\cdot12^2}{2}=-892.5\text{ J}.$$

(b) How much work was done by friction on the block can be found by work-energy theorem: the work done by an external force (friction) equals negative change in (kinetic) energy:

$$W_f=-\Delta KE=892.5\text{ J}.$$