What would be the acceleration of a block sliding down an inclined plane that makes an angle of 45° with the horizontal if the coefficient of sliding friction between two surfaces in 0.3?

Given,

Angle of inclination $(\theta)$ = 45°

Coefficient of Friction $(\mu)$ = 0.3

Downward Acceleration $(a)$ = ?

The downward force $= mg\;Sin\;\theta$

The frictional force = $\mu\;R = \mu\;mg\;Cos\theta$

Net downward force along the plane is,

$F = mg\;Sin\theta - \mu\;mg\;Cos\theta = mg\;(Sin\;45 - 0.3\;Cos\;45) = mg\;*\; 0.495$

Then acceleration $a$ is given by,

$F = ma$     ⇒     $a =$$\frac{F}{m}$ = $\frac{mg\;*\;0.495}{m}$ = $40 * 0.495 = 4.95 \; m/s^2$

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