Question

please explain method and show a picture please

A rope holds a 10-kg rock at rest on a frictionless inclined plane as shown. 30°
Determine the tension in the rope and magnitude of the acceleration of the rock down the inclined plane if the rope breaks.

Answer 1

Here, $mg$ is the weigth of the rock acting downward. R is the normal reaction force to the surface of the inclined plane. as the plane is frictionless, R has no importance in the dynamics of the block.

From the figure it is very clear that as the rock is in equilibrium, the tension in the rope is,

$T = mg \sin \theta = 10\times 9.81\times \sin 30^{\circ} = 49.05 \hspace{2mm} N$

When the rope breakes there is no force to balance the force component $mg \sin \theta$,  therefore the rock start to slide down the plane with that force component. So, in that case the acceleration of the rock is,

$f = \frac{mg \sin \theta}{m} = g \sin \theta = 9.81 \times \sin 30^{\circ } = 4.91 \hspace{2mm} m/s^2$