Question

"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.] "

Answer 1

Answer:

0.42 J

Explanation:

The elastic potential energy of a spring is given by:

$U=\frac{1}{2}kx^2$ (1)

where

k is the spring constant

x is the compression/stretching of the spring

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:

$F=kx$

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

$k=\frac{F}{x}=\frac{12.0 N}{0.070 m}=171.4 N/m$

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

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

Answer 2

We know, Applied force(f) = Spring constant (K) * Extension of material (x)
Here, f = 12 N
x = 0.070 m

Substitute their values,   
12 = k * 0.070
k = 12 / 0.070
k = 171.43 N/m

In short, Your Answer would be 171.43 N/m

Hope this helps!