Question

A particular spring does not obey Hooke's law (F(x) = -kx), but rather is described by F(x)=-k x-kax? x is the displacement of the spring from equilibrium and the constants k, and k2 are given below. Randomized Variables ki = 6.9 N/m k2 = 9.9 N/m2 * 50% Part (a) Input an expression for the potential energy of the spring as a function of position.Part (b) What is the magnitude of the restoring force of this spring, in newtons, if it is stretched 26.5 cm from equilibrium?

Answer 1

Answer:

$\displaystyle U=-3.45x^2-3.3x^3$

$F(0.265)=-2.52\ N$

Explanation:

Spring Potential Energy

For any expression of the restoring force F as a function of the displacement from the equilibrium point x, the change of potential energy is equivalent to the work done to move the object from one point to another.

(a) The potential energy is

$\displaystyle U=\int_{0}^{x}F(x).dx$

We have that

$F(x)=-6.9x-9.9x^2$

Thus

$\displaystyle U=\int_{0}^{x}(-6.9x-9.9x^2).dx$

$\displaystyle U=-6.9\frac{x^2}{2}-9.9\frac{x^3}{3}$

$\displaystyle U=-3.45x^2-3.3x^3$

(b) Knowing that x=26.5 cm = 0.265 m

$F(0.265)=-6.9\cdot 0.265-9.9\cdot 0.265^2$

$F(0.265)=-2.52\ N$