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प्रश्न
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.
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उत्तर
\[\text{Given, }\]
\[\text{ Mass of the block, m = 250 g = 0 . 25 kg } , \]
\[\text{ Spring constant, k = 100 N/m }\]
\[\text{ Compression in the string, x = 10 cm = 0 . 1 m}, \]
\[\text{ Acceleration due to gravity, g = 10 m/ s}^2\]
Let the block rises to height h.
Applying law of conservation of energy which says that the total energy should always remain conserved.
\[\frac{1}{2}\text{kx}^2 = \text{mgh}\]
\[ \Rightarrow h = \frac{1}{2}\left( \frac{\text{kx}^2}{\text{mg}} \right)\]
\[ = \frac{100 \times 0 . 01}{2 \times \left( 0 . 250 \right) \times 10}\]
\[ = 0 . 2 \text{m = 20 cm}\]
So, the block rises to 20 cm.
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