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

D) .35

Step-by-step 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)=(2)/(10)\\\Rightarrow (1)/(2)-\mu(√(3))/(2)=0.2\\\Rightarrow \mu=((1)/(2)-0.2)/((√(3))/(2))\\\Rightarrow \mu=(0.3* 2)/(√(3))\\\Rightarrow \mu=0.3464\approx 0.35`

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