Most Important Selected Qs for JEE AdvancedPhysicsLaws of Motion
A small rubber eraser is placed at one edge of a quarter-circle-shaped track of radius R that lies in a vertical plane and has its axis of symmetry vertical (see figure); it is then released. The coefficient of friction between the eraser and the surface of the track is =0.6 .
Options
- AIf the particle slides from A to B work done by frictional force will be 0.6 mgR 2 .
- BIf the particle slides from A to B work done by frictional force will be greater than 0.6 mg R 4 2 .
- CThe particle will never be able to go from A to B .
- DIf is 2 the particle will not begin to slide.
Correct answer
D. If is 2 the particle will not begin to slide.
Step-by-step solution
Underestimate the work done against friction, and compare it with the initial gravitational potential energy of the eraser. First of all, we investigate whether the rubber eraser will start moving at all. It will do so provided that mg mg , i.e. = 45^ =1 . This is clearly the case, since =0.6 . So, the eraser will start moving. The trouble is that the determination of how the normal force acting on the eraser varies with position is difficult. A calculation of the work done against friction can be carried out, to a