Question

A block of mass $m$ slides from rest at a height $H$ on a frictionless inclined plane as shown in the figure. It travels a distance $d$ across a rough horizontal surface with coefficient of kinetic friction $\mu$, and compresses a spring of spring constant $k$ by a distance $x$ before coming to rest momentarily. Then the spring extends and the block travels back attaining a final height of $h$. Then:

• A. $h=H-2\mu (d+x)$
• B. $h=H+2\mu (d-x)$
• C. $h=H-2\mu d+k{x}^2/mg$
• D. $h=H-2\mu (d+x)+k{x}^2/2mg$

Answer 1

Answer: (A)

$h=H-2\mu (d+x)$

Let us split the motion of the body into 3 parts. Let the mass of the body be $m$

$\text{Part 1: Climbing down the inclined plane}$

External Non conservative forces = 0

$K{E}_i=0;P{E}_i=mgH$

$K{E}_f=K;P{E}_f=0$

By conservation of Mechanical energy, we have $K=mgH$

$\text{Part 2: Forward and Reverse motion on Rough Surface}$

External Non conservative force = Friction = $-\mu mg$

Distance travelled = $2\times (d+x)$ (forward and backward)

Work done = $F\Delta x=-2\mu mg(d+x)$

$K{E}_i=mgH;P{E}_i=0$

$K{E}_f=K;P{E}_f=0$

By work energy theorem, we have

$K-mgH=-2\mu mg(d+x)\Rightarrow K=mgH-2\mu mg(d+x)$

$\text{Part 3: Climbing up the inclined plane}$

External Non conservative forces = 0

$K{E}_i=mgH-2\mu mg(d+x);P{E}_i=0$

$K{E}_f=0;P{E}_f=mgh$

By conservation of Mechanical energy, we have $mgh=mgH-2\mu mg(d+x)$

$\Rightarrow h=H-2\mu (d+x)$