Question

a brick of mass 2kg slides down an incline of height 5m and angle 30.if the coefficient of frictionof the incline is 1/2√3 the velocity of the block at bottom of incline is?

Answer 1

Given:

Mass of brick = 2 kg.

Height of inclined plane is 5 m , angle of inclination is 30°.

Coefficient of friction = 1/(2√3)

To find:

Velocity of the brick at the bottom of the incline.

Calculation:

Considering the Free-Body Diagram of the block and incline system , we can say :

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

Cancelling common terms :

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

$= > \: a = g \sin( 30 \degree) - \mu g \cos( 30\degree)$

$= > \: a = g \bigg \{\sin( 30 \degree) - \dfrac{1}{2 \sqrt{3} } \times \cos( 30\degree) \bigg \}$

$= > \: a = g \bigg \{ \dfrac{1}{2} - ( \dfrac{1}{2 \sqrt{3} } \times \dfrac{ \sqrt{3} }{2} ) \bigg \}$

$= > \: a = g \bigg \{ \dfrac{1}{2} - \dfrac{1}{4} \bigg \}$

$= > \: a = \dfrac{g}{4}$

Using trigonometry , we can find out the length of the inclined plane :

$\therefore \: \dfrac{5}{d} = \sin(30 \degree) \\ = > d = 10 \: m$

Applying Equation Of Kinematics :

$\therefore \: {v}^{2} = {u}^{2} + 2ad$

$= > {v}^{2} = {0}^{2} + \{ 2 \times ( \dfrac{g}{4} ) \times 10 \}$

$= > {v}^{2} = 50$

$= > v = 7.07 \: m {s}^{ - 1}$

So , final answer is :

Velocity at bottom of inclined plane is 7.07 m/s.