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 ?

Answer 1

Given,

$m=100g=0.1kg$

$x=5cm=0.05m$

$k=100N/m$

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

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

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

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

$=1.58m/sec$

For the projectile motion,

Downward direction is taken as negative

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

Now, $y=\left(usin\theta \right)t-\frac{1}{2}g{t}^2$

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

$\Rightarrow t=0.63sec,$

So, $x=\left(ucos\theta \right)t$

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