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

Answer:

500 N/m

Explanation:

The potential energy of a spring is equal to

$PE=\frac{1}{2}kx^2$

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

Solving the equation for k, we get

$\begin{gathered} 2PE=kx^2 \\ k=\frac{2PE}{x^2} \end{gathered}$

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

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

Therefore, the spring constant is 500 N/m