Question

A certain spring found not to obey Hooke's law exerts a restoring force $Fx(x) = -ax - \beta x^2$ if it is stretched or compressed, where $\alpha$ = 60.0 N/m and $\beta$ = 18.0 N/m2. The mass of the spring is negligible. (a) Calculate the potential-energy function U($x$) for this spring. Let $U = 0$ when $x = 0$. (b) An object with mass 0.900 kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 m to the right (the $+x$-direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 m to the right of the $x = 0$ equilibrium position?

Name: Problem 59, Chapter 7: University Physics with Modern Physics
Uploaded: 2019-06-30T02:07:41-08:00
Duration: 5 min 30 s
Channel: Ben Nicholson

   A certain spring found not to obey Hooke's law exerts a restoring force $Fx(x) = -ax - \beta x^2$ if it is stretched or compressed, where $\alpha$ = 60.0 N/m and $\beta$ = 18.0 N/m2. The mass of the spring is negligible. (a) Calculate the potential-energy function U($x$) for this spring. Let $U = 0$ when $x = 0$. (b) An object with mass 0.900 kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 m to the right (the $+x$-direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 m to the right of the $x = 0$
equilibrium position?

University Physics with Modern Physics

Hugh D. Young 14th Edition

Chapter 7, Problem 59

Instant Answer

Solved by Expert Ben Nicholson

06/30/2019

Step 1

This is because the force is the negative derivative of the potential energy. Therefore, we have: \[U(x) = -\int_{0}^{x} Fx(x') dx'\] Substituting the given force function into the integral, we get: \[U(x) = -\int_{0}^{x} (-\alpha x' - \beta x'^{2}) dx'\] This Show more…

Show all steps

Thanks for your feedback!

A certain spring found not to obey Hooke's law exerts a restoring force $Fx(x) = -ax - \beta x^2$ if it is stretched or compressed, where $\alpha$ = 60.0 N/m and $\beta$ = 18.0 N/m2. The mass of the spring is negligible. (a) Calculate the potential-energy function U($x$) for this spring. Let $U = 0$ when $x = 0$. (b) An object with mass 0.900 kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 m to the right (the $+x$-direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 m to the right of the $x = 0$ equilibrium position?