Question

In the arrangement shown, the block is released from rest from $x=0$ (when the spring was in its natural length), on a rough incline surface having $\mu =0.5x.$ The maximum elongation in the spring will be,

• A. $2\text{m}$
• B. $2.5\text{m}$
• C. $1\text{m}$
• D. $1.5\text{m}$

Answer 1

The correct option is A $2\text{m}$

Let the maximum elongation in the spring be $x_m$

Using Work-Energy therorem, $W_{net}=\Delta K$

$W_g+W_f+W_N+W_{\text{spring}}=\Delta K$

$W_N=0$

$W_f={\int }_0^{x_m}f.dx$

Also, $f=\mu N=\mu mg\cos \theta$

$\left(50\sin {37}^{\circ }\right)x_m-{\int }_0^{x_m}\left(0.5x\right(50\left)\right(\cos {37}^{\circ }\left)\right)dx+0-\frac{1}{2}k{x}_m^2=\frac{1}{2}m(v_f^2-v_i^2)$

At the point of maximum elongation $v=0$

Hence, $v_f=0$ and $v_i=0$

$30x_m-10x_m^2-5x_m^2=0$

$\Rightarrow 30=10x_m+5x_m$

$\Rightarrow 15x_m=30$

$\Rightarrow x_m=2\text{m}$

Hence, option $\left(C\right)$ is the correct answer.