Solve the following :


A block of mass $5.0 \mathrm{~kg}$ is suspended from the end of a vertical spring which is stretched by $10 \mathrm{~cm}$ under the load of the block. The block is given a sharp impulse from below so that it acquires an upward speed of $2.0 \mathrm{~m} / \mathrm{s}$. How high will it rise? Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$.


$\mathrm{k}=(\mathrm{mg}) / \mathrm{x}=50 / 0.1=500 \mathrm{~N} / \mathrm{m}$

Total energy T.E. $=\frac{1}{2} \mathrm{~m} \mathrm{v}^{2}+\frac{1}{2} \mathrm{kx} \mathrm{x}^{2} \ldots \ldots .$ (1)

T.E. (at height $h)=\frac{1}{2} k(h-x)^{2}+\operatorname{mgh} \ldots .(2)$

Equating 1 and 2

$\frac{1}{2} m v^{2}+\frac{1}{2} k x^{2}=1 / 2 k(h-x)^{2}+m g h$

$\frac{1}{2} \times 5 \times 2^{2}+\frac{1}{2} \times 500 \times 0.1^{2}=\frac{\frac{1}{2}}{2} \times 500\left(h-0.1^{2}\right)+5 \times 10 \times h$

$\mathrm{h}=0.2 \mathrm{~m}$

$\mathrm{h}=20 \mathrm{~cm}$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now