Question

Two blocks of masses $m_1$ and $m_2$ inter connected 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 the function $x$.

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

Answer 1

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


Since the block $m_2$ does not move force acting on $m_2$ do not perform work.
As blocks do not move in vertical direction, hence work done by normal reaction and weight will be zero.
Then, due to the forces $F_{\text{sp}},F$ and $f_k$, work is done on the block.

Let $v=$ speed of $m_1$

Applying $W-E$ theorem,
$W_{\text{sp}}+W_N+W_{\text{gr}}+W_{f_k}+W_{\text{ext}}=\Delta K$
$(-\frac{1}{2}k{x}^2)+\left(0\right)+\left(0\right)-\mu Nx+Fx=\frac{m_1{v}^2}{2}$
where $N=m_{1}g$

$v=\sqrt{2x(\frac{F}{m_1}-\mu g)-\frac{k}{m_1}{x}^2}$

Hence option A is the correct answer