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Standard XII

Physics

Law of Conservation of Mechanical Energy

A small block...

Question

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

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Solution

Given,
$m = 100 g = 0.1 k g$

$x = 5 c m = 0.05 m$

$k = 100 N / m^{'}$

When the body leaves the spring, let the velocity v.

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

$\Rightarrow v = x \sqrt{\frac{k}{m}}$

$= (0.05) \times \sqrt{\frac{(100)}{(0.1)}}$

$= 1.58 m / s e c$

For the projectile motion,
Downward direction is taken as negative

$θ = 0^{\circ}, y = - 2$

Now, $y = (u, s i n θ) t - \frac{1}{2} g t^{2}$

$- 2 = (- \frac{1}{2}) \times (9.8) \times t^{2}$

$\Rightarrow t = 0.63 s e c,$

So, $x = (u c o s θ) t$

$= (1.58) \times (0.63) = 1 m$

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