Question

FIGURE EX$15.20$ is a kinetic-energy graph of a mass oscillating on a very long horizontal spring. What is the spring constant?

Short answer

The spring constant is$k=4.0N/m$

Step-by-step solution

Given data.

From the graph:

$$\begin{gathered} x_{max}=A=2m \\ {K}_{max}=8.0J \end{gathered}$$

Required: spring constant $k$

Conservation law.

To get a spring constant we will use conservation law which tells that:

1. At max position the potential energy is maxed and kinetic energy is $0$.

2. At equilibrium position the kinetic energy is max and potential energy is $0$.

3. The max kinetic energy = The max potential energy.

According to these notes get that

$U_{max}=K_{max}\Rightarrow \frac{1}{2}k{A}^2=K_{max}$ $\Rightarrow \left(1\right)$

From $\left(1\right)$we get that

$k=\frac{2K_{max}}{A^2}$ $\Rightarrow \left(2\right)$ $\Rightarrow \left(2\right)$

Substitution in $\left(2\right)$to get that

$k=\frac{2\times 8}{{\left(2\right)}^2}=4.0N/m$

$U_{max}=K_{max}\Rightarrow \frac{1}{2}k{A}^2=K_{max}$