Question

a rough inclined plane has angle of inclination 45. A small block is projected up along the plane.If the coefficient of friction is 1/2,the acceleration experienced by the block is(g is acceleration due to gravity)​

Answer 1

Given:

A rough inclined plane has angle of inclination 45°. A small block is projected up along the plane. The coefficient of friction is 1/2.

To find:

Acceleration experienced by block?

Calculation:

Downward acceleration experienced by block will be sum of component of gravity and frictional force.

$\therefore \: F = mg \sin( \theta) + \mu N$

$\implies\: F = mg \sin( \theta) + \mu \{mg \cos( \theta) \}$

$\implies\: ma= mg \sin( \theta) + \mu \{mg \cos( \theta) \}$

$\implies\: a= g \sin( \theta) + \mu \{g \cos( \theta) \}$

$\implies\: a= g \bigg \{ \sin( \theta) + \mu \cos( \theta) \bigg\}$

$\implies\: a= g \bigg \{ \sin( {45}^{ \circ} ) + \dfrac{1}{2} \cos( {45}^{ \circ} ) \bigg\}$

$\implies\: a= g \bigg \{ \dfrac{1}{ \sqrt{2} } + ( \dfrac{1}{2} \times \dfrac{1}{ \sqrt{2} } ) \bigg\}$

$\implies\: a= \dfrac{g}{ \sqrt{2} } \bigg \{ \dfrac{1}{2} + 1 \bigg \}$

$\implies\: a= \dfrac{g}{ \sqrt{2} } \times \dfrac{3}{2}$

$\implies\: a=10.63 \: m {s}^{ - 2}$

So, acceleration is 10.63 m/s².