First slide

Friction on inclined plane of angle more than angle of repose

Question

A block slides down a slope of angle θ with constant velocity. It is then projected up with a velocity of 10 $m s^{- 1}$ , g =10 $m s^{- 2}$ & θ = 30º. The maximum distance it can go up the plane before coming to stop is

Moderate

A

10m
B

5m
C

4m
D

15m

Solution

Since the block slides down the inclined plane with constant velocity, mgsinθ = fk =µk.mgcosθ. During ascent, retardation (a) of the block is a = $g (\sin θ + μ_{k} \cos θ) = 2 g \sin θ$
Now, 0 = $10^{2}$ - 2 x 2gsinθ x s ⇒ S = $\frac{100}{4 g \sin 30^{0}}$ = 5m

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