Question

The expression for downward acceleration of a body on an inclined plane of coefficient of friction m and q inclination is

Answer 1

Answer:

The acceleration of the body down a rough inclined plane is always less than the acceleration due to gravity g. that is a < g. Thus, the minimum force required to push the body down the inclined plane is f (down) = mg( sin Ɵ + μ cos Ɵ )

Explanation:

When you know that F = ma, you can solve for the acceleration. After you solve for the force along the ramp, you can get the acceleration (a = F/m) along the ramp.

Hope this can help you ..

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Answer 2

Acceleration on inclined plane :

• Let's assume that mass 'm' is sliding down on an inclined plane (angle is $\theta$) with coefficient of friction being $\mu$.

Now, we know that :

• Normal force on plane is N = $mg\cos(\theta)$

• Let acceleration be 'a'

$\therefore \: a = \dfrac{mg \sin( \theta) - f }{m}$

$\implies \: a = \dfrac{mg \sin( \theta) - \mu mg \cos( \theta) }{m}$

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

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

This is the expression of acceleration on inclined plane.