First slide

Friction on inclined plane of angle more than angle of repose

Question

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

Moderate

A

1/4
B

1/2
C

3/4
D

√3/2

Solution

For a frictionless surface, o =g sin 45° = g/√2
$∴ l = \frac{1}{2} \times \frac{g}{\sqrt{2}} \times t_{2}^{2}$ …(1)
In presence of friction.
$a = \frac{g}{\sqrt{2}} - \frac{μg}{\sqrt{2}} = \frac{g}{\sqrt{2}} (1 - μ)$
$∴ l = \frac{1}{2} \times [\frac{g}{\sqrt{2}} (1 - μ)] \times t_{1}^{2}$ …(2)
Dividing eq. (2) by eq. (1), we get
$1 = (1 - μ) \frac{{t_{1}}^{2}}{{t_{2}}^{2}} = (1 - μ) \frac{{(2 t_{2})}^{2}}{{t_{2}}^{2}} = (1 - μ) \times 4$
$or (1 - μ) = 1 / 4 or μ = 3 / 4$ .

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