Question

Two blocks of masses $m_1$ and $m_2$ interconnected by a spring of stiffness $k$ are placed on a horizontal surface. If a constant horizontal force $F$ acts on the block $m_1$ it slides through a distance $x$ whereas $m_2$ remains stationary. If the coefficient of friction between all contacting surfaces is $\mu$, find the speed of the block $m_1$ as a function of $x$.

• A. $v=\sqrt{(\frac{F}{m_1}-2\mu g)x-\frac{k}{m_1}{x}^2}$
• B. $v=\sqrt{2(\frac{F}{m_1}-\mu g)x-\frac{2k}{m_1}{x}^2}$
• C. $v=\sqrt{(\frac{F}{m_1}-\mu g)x-\frac{k}{m_1}{x}^2}$
• D. $v=\sqrt{2(\frac{F}{m_1}-\mu g)x-\frac{k}{m_1}{x}^2}$

Answer 1

The correct option is D $v=\sqrt{2(\frac{F}{m_1}-\mu g)x-\frac{k}{m_1}{x}^2}$


Since the block $m_2$ does not move, force acting on $m_2$ does not perform any work. Also as blocks do not move in vertical direction, work done by normal reaction and weight will be zero.
Therefore, spring force ($F_{sp}$), external force($F$) and force due to friction ($f_k$) are only doing work on the block.

Let $v$ = speed of $m_1$ at distance $x$.
Applying $W-E$ theorem,
$W_{sp}+W_N+W_{gr}+W_{fk}+W_{ext}=\Delta K$
$(-\frac{1}{2}k{x}^2)+\left(0\right)+\left(0\right)-\mu N_{1}x+Fx=\frac{m_1{v}^2}{2}$
where $N=m_{1}g$
$\Rightarrow v^2=2(\frac{F}{m_1}-\mu g)x-\frac{k}{m_1}{x}^2$
$\Rightarrow v=\sqrt{2(\frac{F}{m_1}-\mu g)x-\frac{k}{m_1}{x}^2}$