Answer to Question #105042 in Mechanics | Relativity for sridhar

Question

A disc of mass 100 gram slides down from rest on an inclined plane of 30°and comes to rest after travelling a distance of 1m along the horizontal plane.If the coefficient of friction is 0.2 for both inclined and horizontal planes then the work done by the frictional force over the whole journey approximately is (Acceleration due to gravity =10m/s² )
Ans 0.306j

Answer 1

Initially, the disc had potential energy. This energy was wasted to achieve speed and overcome friction, i.e. to gain kinetic energy at the bottom of the incline and work done against friction. Then the kinetic energy (potential energy minus the work done against friction) was wasted fully converted to the work done against friction asking the horizontal. Write this mathematically:

$$mgl\text{ sin}30°=\mu mgl\text{ cos}30°+KE,\\ KE=\mu mgd.$$

Here l - length of the incline, d - horizontal path.

$$mgl\text{ sin}30°=\mu mgl\text{ cos}30°+\mu mgd,\\ \space\\ l=\frac{\mu d}{(\text{sin}30°-\mu \text{ cos}30°)}=0.612\text{ m}.$$

The work done against friction is, therefore, equal to the initial potential energy:

$$W=mgl\text{ sin}30°=0.306\text{ J}.$$