Question

A force of 20 N holds an ideal spring with a 10-N/m spring constant in compression. The potential energy stored in the spring is: a. 0.5 J b. 2.5 J c. 5J d. 20 J e. 200 J

Answer 1

d. 20J

Step-by-step explanation:

An ideal spring experiments a elongation 'x' with a force 'F' according to its spring constant 'k', as:

`F=kx`

If a force of 20 N holds the spring, the spring is exerting a force of -20 N (the spring is flat); it is possible to calculate the elongation:

`x=(-20N)/(10(N)/(m))=-2m`

The potential energy stored is the potential work that the spring could do if the external force disappear, so, according to the differential definition of work,

`dw=Fdx` which means that a differential of work is given by the multiplication of a force F by the displacement dx. This equation can be integrated to give:

`dw=kxdx`

`W=(1)/(2)kx^2`

So, the potential energy is:

`PE=W=(1)/(2)10(N)/(m)*(2m)^2=20J`