Question

a block slides down a rough inclined plane of slope angle theta with a consrant velocity it is projected up the same plane with an initial velocity v the distance travelled by the block up the plane before coming to rest is​

Answer 1

Answer: $d=\frac{v^2}{2g(kcos\alpha+sin\alpha)  }$

(Here, k denotes the coefficient of friction and I've replaced theta by α for typing convenience)

Explanation: We have the following information available with us:

1) Initial velocity= v

2) Final velocity = 0(When the block comes to rest while going up)

3) A rough plane inclined at an angle α with the horizontal.

Assumption: Since no information has been provided regarding the coefficient of friction(which is absurd since the surface is rough), I'll assume k to be the coefficient of friction. Further, I assume that k remains constant throughout the blocks' motion.

First, we compute the effective net force acting on the block. Friction always acts in a direction opposite to that of motion and parallel to the surface, and its magnitude is given by

$f=kN$, where N is the Normal reaction. Since the plane is inclined at an angle α with the horizontal, $N=mgcos\alpha$

⇒$f=kmgcos\alpha$

In addition to friction, the sine component of mg also acts parallel to friction, and its magnitude is $mgsin\alpha$

Thus, net force acting on the body is

$F=-mg(kcos\alpha +sin\alpha )$

⇒$a=\frac{F}{m} =-g(kcos\alpha +sin\alpha )$

(The '-' sign indicates that the force is retarding in nature)

From Newton's Third Equation, the distance d covered by the block is

$d=\frac{-(v)^2}{2a} =\frac{v^2}{2g(kcos\alpha+sin\alpha)  }$

, which is the required expression.