Question

A block slides down a rough inclined plane with a constant velocity \(9.8\,\text{m s}^{-1}\). The coefficient of kinetic friction between the block and the surface is \(\dfrac1{\sqrt3}\). If the block is pushed up along the inclined plane from the bottom with a velocity \(9.8\,\text{m s}^{-1}\), then the distance travelled by the block before coming to rest is

• A. \(9.80\,\text{m}\)
• B. \(2.45\,\text{m}\)
• C. \(14.75\,\text{m}\)
• D. \(4.90\,\text{m}\)

Hint:

If a body moves down an inclined plane with constant velocity, \[ \boxed{\tan\theta=\mu.} \] When moving upward, both gravity and friction oppose the motion, so \[ \boxed{a=g\sin\theta+\mu g\cos\theta.} \]

Answer 1

The Correct Option is D

Step 1: Determine the inclination of the plane. Since the block slides down with constant velocity, \[ mg\sin\theta=\mu mg\cos\theta. \] Thus, \[ \tan\theta=\mu=\frac1{\sqrt3}. \] Hence, \[ \boxed{\theta=30^\circ.} \]

Step 2: Find the retardation while moving upward. While the block moves upward, both gravity and friction act down the plane. Therefore, \[ a=g\sin30^\circ+\mu g\cos30^\circ. \] Now, \[ g\sin30^\circ=\frac{g}{2}, \] and \[ \mu g\cos30^\circ = \frac1{\sqrt3}\times g\times\frac{\sqrt3}{2} = \frac{g}{2}. \] Hence, \[ a=g=9.8\,\text{m s}^{-2}. \]

Step 3: Use the equation of motion. Using \[ v^2=u^2-2as, \] where \[ v=0, \qquad u=9.8\,\text{m s}^{-1}, \qquad a=9.8\,\text{m s}^{-2}, \] we obtain \[ 0=(9.8)^2-2(9.8)s. \] Therefore, \[ s=\frac{9.8}{2}=4.9\,\text{m}. \] Hence, \[ \boxed{s=4.90\,\text{m}.} \] Therefore, the correct option is \(\boxed{(D)}\).