physics question

Question

A 4.00 kg block is released up a 30.0º incline with 128 J of kinetic energy. If the block stops 4.30m up the incline what is the friction force acting on the block? Ans: 10.2N

Answer 1

When the block stops, it will have potential energy equal to height x mass x acceleration of gravity. The height can be obtained by trigonometry as 4.3 m x sin30 = 2.15 m high. The potential energy is 2.15 m x 4.00 kg x 9.8 ms^-2 = 84.28 j. The difference between this PE and the initial KE is 128j - 84.28j = 43.72 j which is the energy lost to overcoming friction.

The work (energy) done against the frictional force is given by force x distance. We know the distance travelled by the block is 4.30 m and the work done is 43.7 j, we can calculate the net force required = 43.7 j / 4.30 m = 10.2 N