A block takes twice as much time to slide down a 45o rough inclined pl...
Given:
Angle of inclined plane = 45o
Coefficient of friction = ?
Let the height of the inclined plane be ‘h’ and distance travelled be ‘d’.
Concept:
When a body slides down an inclined plane, it experiences two types of forces:
1. Gravitational force (mg) acting vertically downwards.
2. Normal force (N) acting perpendicular to the plane and counteracting the gravitational force.
When the plane is rough, there is also a force of friction (f) acting opposite to the direction of motion of the block. The frictional force is given by f = µN, where µ is the coefficient of friction.
When the plane is smooth, there is no frictional force acting on the block. Hence, the block slides down the plane with a greater acceleration than in the case of a rough plane.
Calculation:
Let the time taken to slide down the smooth plane be ‘t’.
Acceleration of the block on the smooth plane, a = g sin 45o
Distance travelled, d = 1/2 a t^2
Time taken, t = √(2d/a)
Let the time taken to slide down the rough plane be ‘2t’.
Acceleration of the block on the rough plane, a' = g sin 45o - µg cos 45o
Distance travelled, d = 1/2 a' (2t)^2 = 2a' t^2
Time taken, 2t = √(2d/a')
As per the given condition, 2t = 2t x (3/2)
=> √(2d/a') = (3/2) √(2d/a)
=> a'/a = 9/4
=> g sin 45o - µg cos 45o = (9/4) g sin 45o
=> µ = (3/4)
Therefore, the coefficient of friction is 3/4. Answer: (a).