Question:
A block of mass $250 \mathrm{~g}$ is kept on a vertical spring of spring constant $100 \mathrm{~N} / \mathrm{m}$ fixed from below. The spring is now compressed to have a length $10 \mathrm{~cm}$ shorter than its natural length and the system is released from this position. How high does the block rise? Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$.
Solution:
By law of conservation of energy:
$\frac{1}{2} \mathrm{kx}^{2}=\mathrm{mgh}$
$\frac{1}{2} \times 100 \times 0.1=\mathrm{mgh}=0.25 \times 10 \times \mathrm{h}$
$h=0.2 \mathrm{~m}$
$\mathrm{~h}=20 \mathrm{~cm}$
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.