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Solve the following :

Question:

A small block of mass $100 \mathrm{~g}$ is pressed against a horizontal spring fixed at one end to compress the spring through $5.0 \mathrm{~cm}$ (figure 8-E11). The spring constant is $100 \mathrm{~N} / \mathrm{m}$. When released, the block moves horizontally till it leaves the spring. Where will it hit the ground $2 \mathrm{~m}$ below the spring?

Solution:

By law of conservation of energy:

$\frac{1}{2} m v^{2}=\frac{1}{2} k x^{2}$

$v=1.58 \mathrm{~m} / \mathrm{s}^{2}$

Now, projectile motion

$y=(u \sin \theta) t-\frac{1}{2} g t^{2}$

$\theta=0^{0}, y=-2$

$t=0.63 \mathrm{sec}$

and

$x=(u \cos \theta) t=1.58 \times 0.63$

$x=1 \mathrm{~m}$

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