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Question

Starting from rest ,a body slides down a 45° inclined plane in twice the time takes to slide down the same distance in the absence of friction .The coefficient of friction between the body and inclined plane is


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Solution

Step 1. Given data.

  1. The inclination of the plane is 45°
  2. The body slide down in twice the time it takes to slide down the same distance in the absence of friction.

Step. 2 Formula used

  1. F=ma
  2. x=ut+12at2

Where F is the force, mthe mass of the body, a the acceleration, x the distance covered, t the time, u the initial velocity and a the acceleration

Step 3. Find the time when there is no friction

Let the acceleration be a1

From newton's laws of motion

F=ma

mgsinθ=ma1 [Where g is the acceleration due to gravity]

a1=gsinθ

Let the body travel x distance in time t1 with the initial velocity u=0m/s

x=ut+12at2x=12a1t12t1=2xa1t1=2xgsinθ

Step 4. Find the time when there is friction

Let the acceleration be a2 and μ the coefficient of friction.

From newton laws of motion

F=ma

mgsinθ-μmgcosθ=ma2 [Where g is the acceleration due to gravity]

a2=gsinθ-μgcosθ

Let the body travel x distance in time t2 with the initial velocity u=0m/s

x=ut+12at2x=12a2t22t2=2xa2t2=2xgsinθ-μgcosθ

Step 5. Find the coefficient of friction

Given that the body slide down in twice the time takes to slide down the same distance in the absence of friction.

2t1=t222xgsinθ=2xgsinθ-μgcosθ8xgsinθ=2xgsinθ-μgcosθ4sinθ=1sinθ-μcosθ

42=21-μ [Substitute the value of θ=45°]

μ=34=0.75

Hence, the coefficient of the friction is 0.75.


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