Question

A body slides down an inclined plane of angle of inclination 30\(^\circ\) with a constant velocity of 10 ms\(^{-1}\). If the body is pushed up the same plane with a velocity of 20 ms\(^{-1}\), the distance moved by the body before coming to rest is (acceleration due to gravity = 10 ms\(^2\))

• A. 10 m
• B. 15 m
• C. 20 m
• D. 30 m

Hint:

When dealing with inclined planes, always consider the components of gravitational force along and perpendicular to the plane. Friction plays a crucial role in determining the motion of the body.

Answer 1

The Correct Option is C

Given: Angle of inclination, \(\theta = 30^\circ\)
Constant velocity while sliding down, \(v = 10 \, {ms}^{-1}\)
Initial velocity while pushing up, \(u = 20 \, {ms}^{-1}\)
Acceleration due to gravity, \(g = 10 \, {ms}^{-2}\) Step 1: Determine the acceleration while sliding down Since the body slides down with a constant velocity, the net acceleration along the incline is zero. This implies that the component of gravitational force along the incline is balanced by the frictional force. \[ mg \sin \theta = \mu mg \cos \theta \] \[ \tan \theta = \mu \] \[ \mu = \tan 30^\circ = \frac{1}{\sqrt{3}} \] Step 2: Determine the acceleration while pushing up When the body is pushed up the incline, the net acceleration \(a\) is given by: \[ a = g \sin \theta + \mu g \cos \theta \] Substituting the values: \[ a = 10 \sin 30^\circ + \frac{1}{\sqrt{3}} \times 10 \cos 30^\circ \] \[ a = 10 \times \frac{1}{2} + \frac{1}{\sqrt{3}} \times 10 \times \frac{\sqrt{3}}{2} \] \[ a = 5 + 5 = 10 \, {ms}^{-2} \] Step 3: Calculate the distance moved before coming to rest Using the equation of motion: \[ v^2 = u^2 + 2as \] Where: - Final velocity, \(v = 0 \, {ms}^{-1}\) - Initial velocity, \(u = 20 \, {ms}^{-1}\) - Acceleration, \(a = -10 \, {ms}^{-2}\) (negative because it is decelerating) \[ 0 = (20)^2 + 2(-10)s \] \[ 0 = 400 - 20s \] \[ 20s = 400 \] \[ s = \frac{400}{20} = 20 \, {m} \] Final Answer:
20 m