for-the-spring-potential-ux-12kx2-with-k-180-nm-what-is-the-force-at-x-020-m

Question

For the spring potential U(x)=12kx2\displaystyle U(x)=\$$frac12kx^2$$ with k=180k = 180  N/m, what is the force at x=−0.20x = -0.20  m?

Accepted Answer

3636  N

Answer

−36-36  N

Answer

−18-18  N

Answer

1818  N

For the spring potential U(x)=12kx2\(\displaystyle\) U(x)=\(\frac\)12kx^2 with k=180k = 180 N/m, what is the force at x=−0.20x = -0.20 m?