Question

Starting from rest, a body slides down a 45° inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is
(a) 0.80 (b) 0.75 (c) 0.25 (d) 0.33

Answer 1

a1 = m g sin x / m = g sin x      acceleration with no friction

a2 = (m g sin x - u m g cos x) / m = g (sin x - u cos x)

S = 1/2 a1 * t1^2 = 1/2 a2 * t2^2 = 2 a2 * t1^2    since t2 = 2 t1

then a1 = 4 a2

g sin x = 4 g (sin x - u cos x)

3 sin x = 4 u cos x

u = 3 sin x / (4 cos x) = 3 tan x / 4

u = .75    since x = 45 deg and tan 45 = 1

Answer 2

Answer:

Explanation:

We know that s = ut + 1/2at2

Now the distance covered is same and the initial velocity is zero, hence by equating the equations for with and without friction we will get

without frictions = 1/2at2 with friction s = 1/2aftf2tf = 2twe get

a = 4af without frictions = 1/2at2

with friction s = 1/2aftf2tf = 2t

we get a = 4af

Acceleration(a) when there is no friction = g Sin 45

Acceleration(af) when there is friction = g Sin 45 - ų g Cos 45

so ​ gSin45 = 4(​g Sin 45 - ų g Cos 45)

or 1 = 4 - 4​ų

Hence ​ų = 0.75