Question

Calculate the elastic potential energy when a 40N force stretches a spring by 0.50cm

Answer 1

Approximately
`0.10\; {\rm J}`, assuming that this spring is an ideal spring.

Step-by-step explanation:

Assume that this spring is ideal. Given the force required to stretch the spring to a specific position, the elastic potential energy stored in that spring at this moment can be found in the following steps:

• Apply Hooke's Law to find the spring constant.
• Obtain an expression for the elastic potential energy in the spring from the spring constant and the displacement of the spring.

By Hooke's Law, the spring constant (force constant)
`k` of an ideal spring is:

`\displaystyle k = -(F)/(x)`,

Where:

• 
  `F` is the restoring force from the spring, and
• 
  `x` is the displacement of the spring.

In this question, it is given that the external force on the spring is
`40\; {\rm N}`. The restoring force from the spring is the reaction force to this external force and would be equal to the opposite of this external force:
`F = (-40)\; {\rm N}` (negative because this force is in the opposite direction.)

It is also given that the displacement of this spring is
`x = 0.50\; {\rm cm}` from the equilibrium position. Apply unit conversion and ensure that all quantities are measured standard units:

`\begin{aligned}x &= 0.50\; {\rm cm} \\ &= 0.50\; {\rm cm}* \frac{1\; {\rm m}}{100\; {\rm cm}} = 5.0* 10^(-3)\; {\rm m}\end{aligned}`.

For an ideal spring with spring constant
`k`, if
`x` is the displacement of the spring from equilibrium position, the elastic potential energy (
`\text{EPE}`) in this spring would be:

`\displaystyle (\text{EPE}) = (1)/(2)\, k\, x^(2)`.

Substitute the Hooke's Law expression for the spring constant
`k = (F/x)` into the expression for
`\text{EPE}` to obtain:

`\begin{aligned} (\text{EPE}) &= (1)/(2)\, k\, x^(2) \\ &= (1)/(2)\, \left((F)/(x)\right)\, x^(2) \\ &= (1)/(2)\, F\, x\end{aligned}`.

Substitute in the value of
`F` and
`x` to find the value of
`(\text{EPE})`:

`\begin{aligned} (\text{EPE}) &= (1)/(2)\, F\, x \\ &= (1)/(2)\, (40\; {\rm N})\, (5.0 * 10^(-3)\; {\rm m}) \\ &= 0.10\; {\rm J}\end{aligned}`.

In other words, under the assumption that this spring is ideal, the elastic potential energy stored in this spring would be
`0.10\; {\rm J}`.