91Ó°ÊÓ

Mass is defined as the object's quantity of inertia or the proportion between the force and acceleration.

When a mass is hung from a vertical spring, the spring expands beyond its normal length. The length change is equivalent to. The gravitational potential energy is equal to zero at the equilibrium point, according to the job. So, the total energy at the equilibrium point is .

$E_{eq}=\frac{k{\left(∆L\right)}^2}{2}$

Let us consider the figure,

As, the gravitational and elastic forces are equivalent at equilibrium point. Write the equation.

$-k∆y=mg$

Substitute $∆y=-∆L$into the obtained equation.

$$\begin{gathered} k∆L=mg \\ ∆L=\frac{mg}{k} \end{gathered}$$

Add all the energies at point yto get the total energy of the system.

$$\begin{gathered} E=\frac{m{v}^2}{2}+\frac{k{\left(y-∆L\right)}^2}{2}+mgy \\ =\frac{m{v}^2}{2}+\frac{k}{2}\left[y^2-2y∆L+{\left(∆L\right)}^2\right]+mgy \\ =\frac{m{v}^2}{2}+\frac{k{y}^2}{2}-ky∆L+\frac{k{\left(∆L\right)}^2}{2}+mgy \end{gathered}$$

Here$E$ is the total energy of system,$m$ is the mass of spring,$K$ is the spring constant,$∆L$ is the change in configuration of length and g is acceleration due to gravity.

Substitute the obtained value of.

$$\begin{gathered} E=\frac{m{v}^2}{2}+\frac{k{y}^2}{2}-ky×\frac{mg}{k}+\frac{k{\left(∆L\right)}^2}{2}+mgy \\ =\frac{m{v}^2}{2}+\frac{k{y}^2}{2}+\frac{k{\left(∆L\right)}^2}{2} \end{gathered}$$

Subtract the energy at equilibrium point from the total energy to get the change in energy,

$$\begin{gathered} ∆E=E-E_{eq} \\ ∆E=\frac{m{v}^2}{2}+\frac{k{y}^2}{2}+\frac{k{\left(∆L\right)}^2}{2}+\frac{k{\left(∆L\right)}^2}{2} \\ =\frac{m{v}^2}{2}+\frac{k{y}^2}{2} \end{gathered}$$

Form the obtained result it is concluded that the horizontal mass-spring system is equal to change in energy.

Therefore, In the horizontal mass-spring system, the change in energy between a point at height and the equilibrium point is$∆E=\frac{m{v}^2}{2}+\frac{k{y}^2}{2}$ .