A certain spring is found not to obey Hooke's law; it exerts a restoring force F (x) equals...


A certain spring is found not to obey Hooke's law; it exerts a restoring force {eq}F_{x} (x) = -\alpha x - \beta x^2 {/eq} , if it is stretched or compressed, where {eq}\alpha {/eq} = 60.0 N/m and {eq}\beta = 18.0 N/m^2 {/eq}.

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 0.9 kg on a friction-less, horizontal surface, is attached to this spring, pulled a distance 1 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 ?

Hooke's Law

Hooke's law states that the force needed to compress or stretch a string is proportional to the amount by which it is compressed or stretched. This is a linear approximation and works only for small stretch and compression. If the spring is stretched or compressed more it doesn't obey Hooke's law and non-linear terms are to be added in the force expression. The non-linear terms make it difficult to find a general equation of motion obeyed by the mass attached to such a spring.

Answer and Explanation:

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{eq}\displaystyle { F_{x} (x) = -\alpha x - \beta x^2 \\ \alpha = 60.0 \ N/m \\ \beta = 18.0 \ N/m^2} {/eq}




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Learn more about this topic:

Hooke's Law & the Spring Constant: Definition & Equation


Chapter 4 / Lesson 19

After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.

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