Question

A block of mass 20kg slides downward a plane inclined at 60° with the horizontal. The coefficient friction between the plane and the block is 0.4. Find the acceleration of the block

Answer 1

The forces acting on the block are shown on the picture. Along the axis, perpendicular to the surface of the plane, $N = m g \cos \theta$.

According to 2nd Newton's law, the acceleration is $m a = m g \sin \theta - f$, where $f$ is the force of friction, $f = \mu N$, which is equal to $f = \mu m g \cos \theta$, using equation above.

Hence,

$m a = m g \sin \theta - \mu m g \cos \theta = m g(\sin \theta - \mu \cos \theta)$, from where the acceleration is $a = g (\sin \theta - \mu \cos \theta) \approx 6.53 \frac{m}{s^2}$.