Question

Starting from rest a body slides down a ${45}^{\circ }$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The co-efficient of friction between the body and the inclined plane is.

• A. 0.75
• B. 0.33
• C. 0.25
• D. 0.8

Answer 1

The correct option is A 0.75

Answer is A.
The normal force, N = $mgsin45$
The friction force, f = $\mu N$
The weight force, W = $mgcos45$ (Not weight of object, but weight force along incline)
The net force F = W-f
Let F1 = Force with no friction, thus F1 = W
Let F2 = Force with friction, thus F2 = W-f
a1 = F1/m
a2 = F2/m
$s=V_{0}t+\frac{1}{2}a{t}^2$
Assume $V_0=0$ (although it subtracts out anyway if it is equal to something)
Since s is the same for both situations and $t2=2\times t1$(as given), $\frac{1}{2}\left(a1\right)(t1{)}^2=\frac{1}{2}(a2\left)\right(2\times t1{)}^2$
(a1)/(a2) = 4
F1/F2 = 4
W/(W-f) = 4
W-f = W/4
(3/4)W = f
0.75($mgcos45$) = $\mu$($mgsin45$)
$\mu$ = 0.75
Hence, the co-efficient of friction between the body and the inclined plane is 0.75