"A jack-in-the-box is a toy in which a figure in an open box is pushed down, compressing a spring. The lid of the box is then closed. When the box is opened, the figure is pushed up by the spring. The spring in the toy is compressed 0.070 meter by using a downward force of 12.0 newtons.

Calculate the total amount of elastic potential energy stored in the spring when it is compressed. [Show all work, including the equation and substitution with units.] "

In order to find the elastic potential energy, we must find the spring constant first. We know by Hook's law that the force compressing the spring is:

[tex]F=kx[/tex]

Re-arranging the equation and substituting F=12.0 N and x=0.070 m, we find:

[tex]k=\frac{F}{x}=\frac{12.0 N}{0.070 m}=171.4 N/m[/tex]

And now we can use eq.(1) to calculate the elastic potential energy:

[tex]U=\frac{1}{2}(171.4 N/m)(0.070 m)^2=0.42 J[/tex]

snehashish65 snehashish65 · Answer 2 · 2022-02-11T19:02:27+01:00

The total amount of elastic potential energy stored in the spring during its compression is 0.42 J.

What is Spring Potential Energy?

When a spring is stretched to some specific distance, then the potential energy stored within the spring is known as spring potential energy or elastic potential energy.

Given data -

The compressed distance of spring is, x = 0.070 m.

The magnitude of downward force is, F = 12.0 N.

The mathematical expression for the elastic potential energy is,

[tex]U = \dfrac{1}{2}kx^{2}[/tex]

Here, k is the spring constant.

Now, using the expression for the spring force as,

[tex]F = k \times x\\\\12.0 = k \times 0.070\\\\k= 12/0.070\\\\k = 171.4 \;\rm N/m[/tex]

Then the elastic potential energy is,

[tex]U = \dfrac{1}{2} \times 171.4 \times 0.070^{2}\\\\U = 0.42 \;\rm J[/tex]

Thus, we can conclude that the total amount of elastic potential energy stored in the spring during its compression is 0.42 J.

Learn more about the elastic potential energy here:https://brainly.com/question/156316