A small block of mass 0.1 kg is pressed against a horizontal spring fixed at one end to compress the spring through 5.0 cm as shown. The spring constant is 100 N/m. When released the block moves horizontally till it leaves the spring. It will hit the ground 2m below the spring at a horizontal distance of

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Detailed Solution

$\frac{1}{2} (100) {(\frac{5}{100})}^{2} = \frac{1}{2} {mv}^{2} \Rightarrow v = \sqrt{\frac{5}{2}};$

$x = v \sqrt{\frac{2 h}{g}} = 1 m$

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