Question

A block slides down a 30.0 degree inclined plane at an acceleration of 2.0 m/s^2. What is the coefficient of friction between the block and the inclined plane? Use g=10.0 m/s^2
A) .20
B) .25
C) .30
D) .35

Answer 1

Answer:

D) .35

Explanation:

m = Mass of block

g = Acceleration due to gravity = $10\ \text{m/s}^2$

$\theta$ = Angle of inclination = $30^{\circ}$

a = Acceleration of block = $2\ \text{m/s}^2$

$\mu$ = Coefficient of friction between the block and the inclined plane

f = Frictional force = $\mu mg\cos\theta$

As the forces are conserved in the system we have

$mg\sin\theta-f=ma\\\Rightarrow mg\sin\theta-\mu mg\cos\theta=ma\\\Rightarrow g(\sin\theta-\mu\cos\theta)=a\\\Rightarrow \sin30^{\circ}-\mu\cos30^{\circ}=\dfrac{2}{10}\\\Rightarrow \dfrac{1}{2}-\mu\dfrac{\sqrt{3}}{2}=0.2\\\Rightarrow \mu=\dfrac{\dfrac{1}{2}-0.2}{\dfrac{\sqrt{3}}{2}}\\\Rightarrow \mu=\dfrac{0.3\times 2}{\sqrt{3}}\\\Rightarrow \mu=0.3464\approx 0.35$

The coefficient of friction between the block and the inclined plane is 0.35