Question

The spring in a retractable ballpoint pen is 1.8 cm long, with a 260 N/m spring constant. When the pen is retracted, the spring is compressed by 1.0 mm. When you click the button to extend the pen, you compress the spring by an additional 6.0 mm.

How much energy is required to extend the pen?

Answer 1

To solve this problem it is necessary to apply the concepts related to the elastic potential energy and the difference that can exist when there are two compression states.

By definition the potential elastic energy is defined as

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

Where

k = Spring constant

x = Displacement

In our case we have two displacements and compression states therefore

`x_1 = 1*10^(-3)m`

When 6 millimeters is compressed, from the first millimeter that was compressed we can define that from the initial state 7mm was compressed

`x_2 = 7*10^(-3)m`

The change in potential energy can then be expressed as

`\Delta PE = PE_2-PE_1`

`\Delta PE = (1)/(2)xk_2^2-(1)/(2)kx_1^2`

`\Delta PE = (1)/(2)k(x_2^2-x_1^2)`

Replacing with our values

`\Delta PE = (1)/(2)(260)(0.007^2-0.001^2)`

`\Delta PE = 0.00624J`

Therefore the total energy required is 0.00624J