First slide

Friction on inclined plane of angle more than angle of repose

Question

Block of mass 1 kg is placed on a rough incline as shown. The coefficient of friction between block and incline is 0.4. The acceleration of block is (g = 10 ms^–2, √3 = 1.7)

Moderate

A

zero
B

5 ms^-2
C

6.5 ms^-2
D

1.6 ms^-2

Solution

Applied force = $mgsin θ = mg \frac{1}{2} = 0.5 mg$
Maximum frictional force
$\begin{aligned} = μmgcos θ \\ = 0.4 mg \frac{\sqrt{3}}{2} \\ = 0.2 \times 1.7 mg \\ = 0.34 mg \end{aligned}$
acceleration = $\frac{0.5 mg - 0.34 mg}{m} = 0.16 g = 1.6 m / s^{2}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App