A spring with a spring constant of 3200 N/m is initially stretched until the elastic potential energy of the spring is 1.44 J. (U =0 for the relaxed spring.)

What is $\mathbf{∆}\mathit{U}$if the initial stretch is changed to

(a) a stretch of 2.0 cm,

(b) a compression of 2.0 cm, and

(c) a compression of 4.0 cm?

The spring constant, k=3200 n/m

The initial potential energy, $U_i=1.44\mathrm{J}$

Use the concept of elastic potential energy of the spring. Find the potential energies at given displacements and then find the change in potential energies.

Formula:

The potential energy is,

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

Here, U is the potential energy, k is the kinetic energy, and x is the distance.

The potential energy of the spring if the initial stretch changed to x=2.0 cm :

You can find the potential energy at

$x=2.0\mathrm{cm}=0.02\mathrm{m}$

The potential energy is,

$$\begin{gathered} U_1=\frac{1}{2}\left(3200\mathrm{N}/\mathrm{m}\right){\left(0.02\mathrm{m}\right)}^2 \\ =0.64\mathrm{J} \end{gathered}$$

The change in potential energy you get,

$$\begin{gathered} ∆U=U_1-U_i \\ =0.64\mathrm{J}-1.44\mathrm{J} \\ =-0.8\mathrm{J} \end{gathered}$$

Potential energy of the spring if initial compression changed to

x =2.0 cm

The answer will be same as part (a) because the spring is elongated at 2.0 cm . It doesn’t matter whether it is stretched or compressed. Potential energy will be the same.

Potential energy of the spring if initial compression changed to

$x=4.0\mathrm{cm}=0.04\mathrm{m}$

Therefore, the potential energy is define as below.

$$\begin{gathered} U_2=\frac{1}{2}\left(3200\mathrm{N}/\mathrm{m}\right){\left(0.04\mathrm{m}\right)}^2 \\ =2.56\mathrm{J} \end{gathered}$$

Determine the change in potential energy as follow.

$$\begin{gathered} ∆U=U_2-U_i \\ =2.56\mathrm{J}-1.44\mathrm{J} \\ =1.12\mathrm{J} \end{gathered}$$

Hence, the required change in potential energy is $∆U=1.12J$.