# Question #57654

Jun 9, 2015

About $\text{5 N}$ [down the incline].

#### Explanation:

The force acting on the object is equal to the object's weight $m g$

On an incline, this force is divided into two components.

Component 1 directed into the incline = $m g \cos \theta$

This force is balanced by the normal force (the supporting force exerted by the incline on the object)

Component 2 is parallel to the incline = $m g \sin \theta$

This is the unbalanced (net force) on the object--take $g = {\text{10 m s}}^{- 2}$

${F}_{\text{net"=mgsintheta =(2)(g)5/20 = "5 N}}$