91Ó°ÊÓ

1. Mass of stone, $m=8.0kg$
2. Compression of spring,$y=0.100m$

The problem deals with the law of conservation of energy and Hooke’s law. According to the law of energy conservation, energy can neither be created, nor be destroyed.Hooke’s law states that the displacement or size of a deformation is directly proportional to the deforming force or load for relatively modest deformations of an object.Using the energy conservation law and Hooke’s law, find the spring constant, change in potential energy and maximum height reached by the stone after releasing it.

Formula:

Force is given by,

$f=ma$

Force by Hook’s law is given by,

$f=kx$

Elastic potential energy,

$U=\frac{1}{2}k{y}^2$

where, Uis potential energy, m is mass, v is velocity, a is an acceleration, x, y are displacements, k is spring constant, and F is force.

By Hooke’s law,

$F=kx$

And by Newton’s law,

$f=ma$

So,

$$\begin{gathered} kx=mg \\ k\left(0.10\right)=8\left(9.8\right) \\ k=784\text{N/m} \end{gathered}$$

Hence, spring constant is $784N/m$.

If the spring is compressed for 30.0 cm again, the total compression is,

So, as per the elastic potential energy formula,

$$\begin{gathered} U=\frac{1}{2}k{y}^2 \\ U=\frac{1}{2}\left(784\right){\left(0.400\right)}^2 \\ U=\frac{1}{2}\left(784\right){\left(0.400\right)}^2 \\ U=62.7J \end{gathered}$$

Hence, elastic potential energy of the compressed spring just before it released after pulling down is $62.7J$.

As, thestone is released from thecompressed spring, it has potential energy, during movement, it has kinetic energy, and at thehighest point, it has only potential energy as the highest point doesn’t have velocity. So, energy will remain constant. That is,$62.7J$

Hence, change in gravitational potential energy of system is $62.7J$.

At maximum height, the stone has only potential energy. So,

$$\begin{gathered} mgh=62.72J \\ \left(8\right)\left(9.8\right)h=62.72J \\ h=0.8m \end{gathered}$$

Hence, maximum height reached by the stone after release is $0.8m$.