Question

A block of mass m is moving with a speed $v$ on a horizontal rough surface and collides with a horizontally mounted spring of spring constant $k$ as shown in the figure. The coefficient of friction between the block and the floor is $\mu$ The maximum compression of the spring is

• A. $-\frac{\mu \text{mg}}{k}+\frac{1}{k}\sqrt{{\left(\mu \text{mg}\right)}^2+mk{v}^2}$
• B. $\frac{\mu \text{mg}}{k}+\frac{1}{k}\sqrt{{\left(\mu \text{mg}\right)}^2+mk{v}^2}$
• C. $-\frac{\mu \text{mg}}{k}-\frac{1}{k}\sqrt{{\left(\mu \text{mg}\right)}^2-mk{v}^2}$
• D. $\frac{\mu \text{mg}}{k}-\frac{1}{k}\sqrt{{\left(\mu \text{mg}\right)}^2+mk{v}^2}$

Answer 1

Answer: (A)

In presence of friction, both the spring force and the frictional act so as to oppose the compression of the spring.
Work done by the net force
$W=-\frac{1}{2}k{x}_m^2-\mu mg{x}_m$
where $x_m$ is the maximum compression of the spring.
Change in kinetic energy
$\Delta K=K_f-K_i=0-\frac{1}{2}m{v}^2$
According to work-energy theorem
$W=\Delta K$
$-\frac{1}{2}k{x}_m^2-\mu mg{x}_m=-\frac{1}{2}m{v}^2$
$k{x}_m^2+\mu mg{x}_m=\frac{1}{2}m{v}^2$
$k{x}_m^2+2\mu mg{x}_m-m{v}^2=0$
$x_m^2+\frac{2\mu mg{x}_m}{k}-\frac{m{v}^2}{k}=0$
It is a quadratic equation in $x_m$.
Solving this quadratic equation for $x_m$
and taking only the positive root since $x_m$ is positive,
we get
$x_m=-\frac{\mu mg}{k}+\frac{1}{k}\sqrt{(\mu mg{)}^2+mk{v}^2}$