Question #57654

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Answer 1 · 2015-06-09T10:57:15

About $5 N$ $"5 N"$ [down the incline].

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

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

Component 1 directed into the incline = $m g cos θ$ $mgcostheta$

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

![http://www.sparknotes.com/testprep/books/sat2/physics/chapter8section8.rhtml]( useruploads.socratic.org )

Component 2 is parallel to the incline = $m g sin θ$ $mgsintheta$

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

$F_{net} = m g sin θ = (2) (g) \frac{5}{20} = 5 N$ $F_"net"=mgsintheta =(2)(g)5/20 = "5 N"$