Question

A small block of mass 100 g is pressed against a horizontal spring fixed at one end to compress the spring through 5 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?
Figure

Answer 1

It will hit the ground 2m below the spring at horizontal distance 1m away from the free end of the spring

1. When block is released the potential energy converts into kinetic energy
2. By the law of energy conservation
3. $\frac{1}{2} kx^{2} =\frac{1}{2}mV^{2}$
4. here k is the spring constant ,x is the compressed distance of spring, m is the mass of the block and V is the speed of block
5. $V=\sqrt{\frac{kx^{2} }{m} }$
6. Time of flight $T= \sqrt{\frac{2H}{g} }$
7. here H is the height and g is the acceleration due to gravity
8. The horizontal distance travelled by the block is
9. $VXT=\sqrt{\frac{kx^{2} }{m }X\frac{2H}{g}$
10. Horizontal distance = $\sqrt{\frac{100(0.05)^{2} }{0.1 }X\frac{2(2)}{10}$ = 1m

Answer 2

Block will hit the ground 2 m below the spring at a distance of 1m.

Explanation:

From the question, it can be noticed that

m = 100g = 0.1kg,

k = 100N/m

and x = 5cm = 0.05m

when the spring is left by the body.

Let us denote the velocity as v

$1 / 2 m v^{2}=1 / 2 k x^{2}$

         $\mathrm{v}=\mathrm{x} \sqrt{\mathrm{k} / \mathrm{m}}=0.05 \times \sqrt{\frac{100}{0.1}}=1.58 \mathrm{m} / \mathrm{sec}$

From the motion of the projectile,  $\theta=0^{\circ}, Y=-2$

It is now seen that

$\begin{array}{l}y=(u \sin \theta) t-1 / 2 g t^{2} \\-2=(-1 / 2) \times 9.8 \times t^{2} \Rightarrow t=0.63 \mathrm{sec}\end{array}$

Thus, x = (u cos θ)t = 1.58 (0.63) = 1 m.

The block hits the ground 2 m below the spring at x = 1m