Question

4. A spring is stretched 0.5 m from equilibrium. The force constant (k) of the

Answer 1

Answer:

31.25

Explanation:

The formula for the potential energy of a spring is $\frac{1}{2}kx^2$, where x is the distance in meters the spring is stretched. Plugging in the numbers that you are given, you get $\frac{1}{2}\cdot 250\cdot 0.25=31.25$. Hope this helps!

Answer 2

31.25 hope this helps:)