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A block slides down an inclined plane of slope angle θ with constant velocity. It is then projected up the same plane with an initial speed (a) How far up the incline will it move before coming to rest? (b) Will it slide down again?

A block slides down an inclined plane of slope angle θ with constant velocity. It is then projected up the same plane with an initial speed  (a) How far up the incline will it move before coming to rest? (b) Will it slide down again?

Grade:11

1 Answers

Kevin Nash
askIITians Faculty 332 Points
8 years ago
(a)
The block moves down the inclined plane with constant speed, therefore kinetic force of friction is equal to the component of its weight along the length of inclined plane, that is
fk = mg sin θ
Here m is the mass of the block, and g is the acceleration due to gravity.
The frictional force is also given as
234-832_1.PNG
The acceleration experienced by the block when it is projected upwards is:

234-883_1.PNG
It has been calculated above that the coefficient of kinetic friction is μk = tan θ . Substitute the same in the equation above,
234-2218_1.PNG
Therefore the block decelerates at a = - 2g sin θ (negative sign accounts for the fact that the block slows down).
The time taken by the block to come at rest is:
234-831_1.PNG
Also, the distance travelled by the block is:

234-713_1.PNG
(b)
The coefficient of static friction is always greater than the coefficient of kinetic friction, therefore if the block stopped once in its journey it will not going to move again unless the force accelerating the block has a magnitude greater than static force of friction.

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