Question

40. When a spring is stretched 0.200 meter from its equilibrium position, it possesses a potential energy of 10.0 joules. What is the spring constant for this spring?

Answer 1

500 N/m

Step-by-step explanation:

The potential energy of a spring is equal to

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

Where k is the spring constant and x is the distance stretched.

Solving the equation for k, we get

`\begin{gathered} 2PE=kx^2 \\ k=(2PE)/(x^2) \end{gathered}`

Now, we can replace PE = 10.0 J and x = 0.200 m, so

`k=\frac{2(10.0J)}{(0.2\text{ m)}^2}=(20.0J)/(0.04m^2)=500N/m`

Therefore, the spring constant is 500 N/m