A body takes 1 1/3 times as much time to slide down a rough inclined p...
A body takes 1 1/3 times as much time to slide down a rough inclined p...
To solve this problem, we need to consider the forces acting on the body as it slides down the inclined plane. Let's analyze the forces on the body for both the rough and smooth inclined planes separately.
For the smooth inclined plane:
- The only force acting on the body is its weight, given by mg, where m is the mass of the body and g is the acceleration due to gravity.
- The weight can be resolved into two components: mg sinθ, which acts parallel to the incline, and mg cosθ, which acts perpendicular to the incline.
- The force of friction is absent on the smooth inclined plane, as there is no roughness to oppose the motion.
- The net force acting on the body is the component of its weight parallel to the incline, which is mg sinθ.
For the rough inclined plane:
- The forces acting on the body are the weight and the force of friction.
- The weight can be resolved into the same components as before: mg sinθ and mg cosθ.
- The force of friction opposes the motion of the body and acts parallel to the incline.
- The net force acting on the body is the component of its weight parallel to the incline minus the force of friction, which is (mg sinθ) - (μmg cosθ), where μ is the coefficient of friction.
Now, let's consider the given information that the body takes 1 1/3 times as much time to slide down the rough inclined plane as it takes to slide down the smooth inclined plane.
- Let's assume the time taken to slide down the smooth inclined plane is t.
- The time taken to slide down the rough inclined plane is then 4/3 times t.
- Since the distance traveled is the same for both cases, we can equate the distances traveled on each plane.
- The distance traveled on the smooth inclined plane is given by d = (1/2)at², where a is the acceleration.
- The distance traveled on the rough inclined plane is also given by d = (1/2)at², where a is the net acceleration.
- Equating these distances, we have (1/2)at² = (1/2)(μmg cosθ)t².
- Simplifying, we get μ = 1/4.
The coefficient of friction is given by μ = 1/4, which is equivalent to 7/16. Therefore, the correct answer is option A, 7/16.
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