Question

In the position shown in Fig. 8.251, the spring is at its natural length. The block of mass m is given a velocity $v_0$ towards the vertical support at t = 0. The coefficient of friction between the block and the surface is given by $\mu =\alpha x$, where $\alpha$ is a positive constant and x is the position of the block from its starting position. The block comes to rest for the first time at x, which is :

• A. $v_1\sqrt{\frac{m}{k+2\alpha mg}}$
• B. $v_0\sqrt{\frac{m}{k}}$
• C. $v_0\sqrt{\frac{m}{\alpha g}}$
• D. None of these

Answer 1

Answer: (A)

$v_1\sqrt{\frac{m}{k+2\alpha mg}}$

By energy can,

$\Rightarrow \frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2+\alpha xmg\times x$

$\Rightarrow \frac{1}{2}m{v}^2=x^2[\frac{k}{2}+\alpha mg]$

$\Rightarrow x^2\left[\frac{k+2\alpha mg}{2}\right]=\frac{1}{2}m{v}^2$

$\Rightarrow x=v_1\sqrt{\frac{m}{k+2\alpha mg}}.$

Hence, the answer is not given in option.