First slide

Friction on inclined plane of angle more than angle of repose

Question

A block of mass 0 . 5 kg has an initial velocity of 10 m/s down an inclined rough plane of angle 30⁰ as shown in fig. (6). The coefficient of friction between the block and inclined surface being 0 2. The velocity of the block after it travels a distance of 10 m is

Moderate

A

2a m/s
B

13 m/s
C

8 m/s
D

17 m/s

Solution

The acceleration is given by
$a = g (\sin 30^{\circ} - μcos 30^{\circ}) = 9 \cdot 8 (\frac{1}{2} - 0 \cdot 2 \times \frac{\sqrt{3}}{2}) = 9 \cdot 8 \times (0 \cdot 3268) = 3 \cdot 2 m / s^{2} Now v^{2} = u^{2} + 2 as v^{2} = (10)^{2} + 2 \times 3 \cdot 2 \times 10 = 164 v \approx 13 m / s$

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