Question

As shown in figure there is a spring block system.Block of mass 500g is presses against a horizontal spring fixed at on end to compress the spring through 5 cm.The spring constant is 500N/m.When Released,the block moves horizontally till it leaves the spring.Calculate the distance where it will hit the ground 4m below the spring​

Answer 1

$\huge\star{\bold{\underline{\underline{\tt{\red{Answer \implies \sqrt{2}\: metre}}}}}}\star$

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$\large\star{\bold{\underline{\mathscr{\green{Explanation}}}}}\star$

$\bigcirc$When block is released,the block moves with speed V till it leaves the spring.

$\implies$By energy conservation $\frac{1}{2}$kx²=$\frac{1}{2}$mv²

Here,

V²=$\large\frac{k{x}^{2} }{m}$

$\large v = \sqrt{ \frac{k {x}^{2} }{m} }$

We know that,

$\bigcirc$Time of Flight = $\sqrt{\frac{2H}{g}}$

So, horizontal distance travelled from the free end of the spring is V x t

$= \sqrt{ \frac{k {x}^{2} }{m} } \times \sqrt{ \frac{2h}{g} } \\ \\ = \sqrt{ \frac{500 \times {0.05}^{2} }{0.5} } \times \sqrt{ \frac{2 \times 4}{10} } \\ \\ = \sqrt{2}$

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Answer 2

$\huge\bf{Answer:-}$

Refer the attachment.