Question

A small rectangular block starting from rest slides down a rough inclined plane of inclination 37° with the horizontal. It's displacement in meters's is
given as a function of time 't' expressed in seconds by: s = t^2 .The coefficient of friction between the block and the inclined surface is (g=10m/s2 and tan 37° = 3/4)
(A) 1/4
(B) 3/4
(C) 1/root2
(D) 1/2​

Answer 1

Given:

A small rectangular block starting from rest slides down a rough inclined plane of inclination 37° with the horizontal. It's displacement in meters's is given as a function of time 't' expressed in seconds by: s = t².

To find:

Coefficient of friction between block and inclined plane ?

Calculation:

$\therefore \: s = {t}^{2}$

$\implies \:v = \dfrac{ds}{dt}$

$\implies \:v = \dfrac{d( {t}^{2} )}{dt}$

$\implies \:v = 2t$

$\implies \: a = \dfrac{dv}{dt}$

$\implies \: a = \dfrac{d(2t)}{dt}$

$\implies \: a = 2 \: m {s}^{ - 2}$

So, net acceleration of block is 2 m/s².

Now. according to FBD of block :

$\therefore \: mg \sin( \theta) - \mu mg \cos( \theta) = ma$

$\implies\: g \sin( \theta) - \mu g \cos( \theta) = a$

$\implies\: 10 \sin( {37}^{ \circ} ) - \mu (10)\cos( {37}^{ \circ} ) = 2$

$\implies\: \sin( {37}^{ \circ} ) - \mu \cos( {37}^{ \circ} ) = \dfrac{2}{10}$

$\implies\: \sin( {37}^{ \circ} ) - \mu \cos( {37}^{ \circ} ) = \dfrac{1}{5}$

$\implies\: \dfrac{3}{5} - \mu ( \dfrac{4}{5} )= \dfrac{1}{5}$

$\implies\: \dfrac{4 \mu}{5} = \dfrac{3}{5} - \dfrac{1}{5}$

$\implies\: \dfrac{4 \mu}{5} = \dfrac{2}{5}$

$\implies\: \mu = \dfrac{1}{2}$

So, final answer is:

$\boxed{ \bold{\: \mu = \dfrac{1}{2} }}$