Question

A wooden block of mass m = 9 kg starts from rest on an inclined plane sloped at an angle θ from the horizontal. The block is originally located 5m from the bottom of the plane. If the block, undergoing constant acceleration down the ramp, slides to the bottom in t = 2 s, and θ = 30º, what is the magnitude of the kinetic frictional force on the block? ƒk =

Answer 1

Answer:

f = 0.283

Explanation:

With the given values x=5m and t=2s, the acceleration a of the block must be:

(1) $a=\frac{2x}{t^2}$

The sum of all forces in the inclined plane must be:

(2) $F=ma=sin(\theta) F_{gravity}-F_{friction}=sin(\theta) mg - f cos(\theta) mg$

Solving equation2 for the acceleration a:

(3) $a=sin(\theta) g - f cos(\theta) g$

Using equations 1 and 3 to solve for f:

$\frac{2x}{t^2}=sin(\theta) g - f cos(\theta) g\\f=tan(\theta)-\frac{2x}{t^2cos(\theta)g}$