Question

What is the acceleration of a block on a ramp inclined 35o to the horizontal if µk = 0.4?

Answer 1

We can solve the problem by applying Newton's second law, which states that the resultant of the forces acting on an object is equal to the product between its mass and its acceleration:
$\sum F = ma$

We should consider two different directions: the direction perpendicular to the inclined plane and the direction parallel to it. Let's write the equations of the forces along the two directions, decomposing the weight of the object (mg):

$mg \sin \theta - \mu_K N = ma$ (parallel direction) (1)
$mg \cos \theta - N =0$ (perpendicular direction) (2)
where
$\theta=35^{\circ}$ is the angle of the inclined plane, N is the normal reaction of the plane, $\mu_K N$ is the frictional force, with $\mu_K=0.4$ being the coefficient of friction.

From eq.(2), we find
$N=mg \cos \theta$
and if we substitute into eq.(1), we can find the acceleration of the block:
$mg \sin \theta - \mu_k mg \cos \theta = ma$
from which
$a=g(\sin \theta - \mu_K \cos \theta)=(9.81 m/s^2)(\sin 35^{\circ} - 0.4 \cos 35^{\circ})=2.41 m/s^2$