Question

A body of mass 10kg is sliding down a rough inclined plane which makes an angle of 30º with the horizontal.If the co efficient of friction is 0.25.Find the acceleration of the body.

Answer 1

Given :

• mass of the body = 10 kg
• angle of the inclined plane = 30 °
• co efficient of friction = 0.25

Formula :

Acceleration of the body on an inclined plane is given by the formula ,

$\boxed{a = g ( sin \: \theta- \mu \: cos \theta)}$

here ,

• a = acceleration
• $\mu \: = coefficient \: of \: friction$

Derivation :

When an object of mass ( m ) is sliding down an inclined plane the friction force ( F ) is acting in upward direction and sin component of gravity is acting in downward direction

$\implies{mg \: sin \theta}$

And the normal force acting on the object is equal to the cos component of gravity

$\implies{N = mg \: cos \theta}$

we know that ,

$\implies{F = \mu \: N }$

$\implies{F = \mu \: mg \: cos \theta}$

Now applying Newton's second law of motion we get ,

$\implies{ma \: = mgsin \theta - F}$

$\implies{ma \: = mgsin \theta - \mu \: mgcos \theta}$

$\implies{a \: = gsin\theta - \mu \: gcos \theta}$

$\boxed{a \: = g(sin \theta - \mu \: cos \theta)}$

Solution:

Substituting the given values in the above equation we get

$\implies{a \: = 9.8(sin 30 - 0.25 \times \: cos 30)}$

$\implies{a \: = 9.8( \frac{1}{2} - 0.25 \times \frac{1.73}{2} )}$

$\implies{a \: = 9.8( 0.5 - 0.25 \times 0.86)}$

$\implies{a \: = 9.8( 0.5 - 0.215)}$

$\implies{a = 9.8 \times 0.285}$

$\boxed{a = 2.8 \frac{m}{ {s}^{2} }}$