Question

For the block moving as shown in figure, maximum compression in the spring will be approximately:
[Take $g=10{\text{m/s}}^2,\sqrt{1552-40\sqrt{3}}\approx 38.50\&10+2\sqrt{3}\approx 13.5$ for calculation. Neglect the effect of corner]

• A. $2.5\text{m}$
• B. $3\text{m}$
• C. $2.1\text{m}$
• D. $2\text{m}$

Answer 1

The correct option is A $2.5\text{m}$

Let the maximum compression in spring be $x$. In this state, the block is momentarily at rest i.e $v=0$.


Displacement of block along horizontal surface is $d_1=2\text{m}$;
Displacement of block along inclined surface is $d_2=(2+x)\text{m}$
On applying Work Energy theorem on the block from initial position ($u=10\text{m/s}$) to final position ($v=0$),


$W_{\text{all Forces}}=\Delta K.E$
$\Rightarrow W_{\text{friction}}+W_{\text{gravity}}+W_{\text{spring}}=K.E_f-K.E_i$
$\Rightarrow [-\mu mg{d}_1-\mu N(d_2+x\left)\right]-mg\left(\right(d_2+x\left)\sin 30\right)-\frac{1}{2}k{x}^2=0-\frac{1}{2}m{u}^2...\left(i\right)$
Here:
$W_{\text{friction}}=-\mu mg{d}_1-\mu N(d_2+x)$
[$\because N=mg\cos {30}^{\circ }$ on the incline]
$=(-0.2\times 2\times 10\times 2)-(0.2\times 2\times 10\times \frac{\sqrt{3}}{2}\times (2+x\left)\right)$
$W_{\text{gravity}}=-mg(d_2+x)\sin {30}^{\circ }$
$=-(2\times 10\times (2+x)\times \frac{1}{2})$
$W_{\text{spring}}=-\frac{1}{2}k{x}^2=-(\frac{1}{2}\times 10\times x^2)$
$\Delta KE=-\frac{1}{2}\times 2\times {10}^2$
Substituting the values in Eq.$\left(i\right)$,
$\Rightarrow -8-2\sqrt{3}(2+x)-10(2+x)-5x^2=-100$
$\Rightarrow -8-4\sqrt{3}-2\sqrt{3}x-20-10x-5x^2=-100$
$\Rightarrow 5x^2+x(10+2\sqrt{3})-72+4\sqrt{3}=0$
$\Rightarrow x=\frac{-(10+2\sqrt{3})\pm \sqrt{(112+40\sqrt{3}+1440)-80\sqrt{3}}}{10}$
Rejecting the $-ve$ value as $x$ can't be negative,
$\Rightarrow x=\frac{-13.46+\sqrt{1552-40\sqrt{3}}}{10}$
$\Rightarrow x=\frac{-13.46+38.50}{10}$
$\therefore x\simeq 2.50$
Maximum compression in the spring is approximately $2.5\text{m}$.