Question

A box slides down a frictionless plane inclined at an angle θ ¸ above the horizontal. The gravitational force on the box is directeda)vertically.b)perpendicular to the plane.c)at an angle θ ¸ below the inclined plane.d)parallel to the plane in the same direction as the movement of the box.e)parallel to the plane in the opposite direction as the movement of the box.

Answer 1

Answer: at an angle $\theta$ below the inclined plane.

If we draw the Free Body Diagram for this situation (figure attached), taking into account only the gravity force in this case, we will see the weight $W$ of the block, which is directly proportional to the gravity acceleration $g$:  

$W=m.g$

This force is directed vertically at an angle $\theta$ below the inclined plane, this means it has an X-component and a Y-component:

$W=W_{X}+W_{Y}$

$W_{X}=m.g.cos(\theta)$

$W_{Y}=m.g.sin(\theta)$

Therefore the correct option is c

Answer 2

Explanation:

When a box is placed in the friction less plane inclined at an angle θ. The forces acting on the box are

1. Weight of the box, W = m g

2. Force of friction, f

3. Normal force, N

When we resolve these forces in horizontal and vertical components, we get :

$N=mg\ cos\theta$

$f=mg\ sin\theta$

We know that the direction of gravitational force is always in downward direction. So, the gravitational force on the box is directed vertically. Hence, the correct option is (a).