Question

A block of mass $5\text{kg}$ is released from rest when compression in spring is $2\text{m}$. Block is not attached with the spring of spring constant $k$, natural length of the spring is $4\text{m}$. The maximum height of the block from the ground is:
$(g=10{\text{m/s}}^2)$

• A. $5.5\text{m}$
• B. $4.5\text{m}$
• C. $6\text{m}$
• D. $7.5\text{m}$

Answer 1

The correct option is A $5.5\text{m}$

Given:
$m=5\text{kg};x=2\text{m};k=300\text{m/s}$



Since, the surface of wedge is smooth, so we can apply conservation of mechanical energy between initial and final points.

$(PE{)}_i+(KE{)}_i=(PE{)}_f+(KE{)}_f$

Taking ground as a reference,

${[\frac{1}{2}k{x}^2+mg(2\sin {30}^{\circ })+0]}_i={[0+mg(6\sin {30}^{\circ })+\frac{1}{2}m{v}^2]}_f$

${\left[\frac{1}{2}300\right(2{)}^2+50\left(2\sin {30}^{\circ }\right)]}_i={\left[50\right(6\sin {30}^{\circ })+\frac{1}{2}5v^2]}_f$

$\therefore v^2=200.......\left(1\right)$


Now, block will follow the projectile path.


The maximum height of a projectile from the ground surface is given by

$H=3+h_m=3+\frac{v^2{\sin }^2{30}^{\circ }}{2g}$

From equation $\left(1\right)$ equation $v^2=200$

$H=3+2.5=5.5\text{m}$

So, the maximum height of the block from the ground is $5.5\text{m}$.

Therefore, option $\left(A\right)$ is correct.