Question

A block of mass 250 g is kept on a vertical spring of spring constant 100 N/m fixed from below. The spring is now compressed 10 cm shorter than its natural length and the system is released from this position. How high does the block rise? Take g = 10 m/s².

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=250g=0.25\mathrm{kg}, \\ \mathrm{Spring}\mathrm{constant},k=100\mathrm{N}/\mathrm{m} \\ \mathrm{Compression}\mathrm{in}\mathrm{the}\mathrm{string},x=10\mathrm{cm}=0.1\mathrm{m}, \\ \mathrm{Acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity},g=10\mathrm{m}/s^2 \end{gathered}$$
Let the block rises to height h.
Applying law of conservation of energy which says that the total energy should always remain conserved.

$$\begin{gathered} \frac{1}{2}k{x}^2=mgh \\ \Rightarrow h=\frac{1}{2}\left(\frac{k{x}^2}{mg}\right) \\ =\frac{100\times 0.01}{2\times \left(0.250\right)\times 10} \\ =0.2\mathrm{m}=20\mathrm{cm} \end{gathered}$$

So, the block rises to 20 cm.