First slide

Friction

Question

A block, released from rest from the top of a smooth inclined plane of inclination $θ$ , has a speed v when it reaches the bottom. The same block, released from the top of a rough inclined plane of the same inclination $θ$ , has a speed v/n on reaching the bottom, where n is a number greater than unity. The coefficient of friction is given by

Moderate

A

$μ = (1 - \frac{1}{n^{2}}) \tan θ$
B

$μ = (1 - \frac{1}{n^{2}}) \cot θ$
C

$μ = {(1 - \frac{1}{n^{2}})}^{1 / 2} \tan θ$
D

$μ = {(1 - \frac{1}{n^{2}})}^{1 / 2} \cot θ$

Solution

We use the relation $v^{2} - u^{2} = 2 as$ . Since u = 0, we have $v^{2} = 2 as$ .
Now $v_{1}^{2} = 2 a_{1} s or v^{2} = 2 g \sin θ \times s$
and $v_{2}^{2} = 2 a_{2} s or \frac{v^{2}}{n^{2}} = 2 (gsin θ - μgcos θ) \times s$
Dividing, we get
$n^{2} (\sin θ - μcos θ) = \sin θ$
which gives $μ = (1 - \frac{1}{n^{2}}) \tan θ$ , which is choice (1).

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