Chapter 5, Problem 5.1P

(a)

To determine

The condition that determines $l_1$ and to show that the net force on the mass is $F=-kx$.

Answer to Problem 5.1P

The net force acting on the spring is $F_{\text{net}}=-kx$

Explanation of Solution

The stretched length of the spring $l_1$ when the mass is attached to it is determined by the force constant $k$ of the spring and the mass $m$ that is hung to it. These are the conditions that determent stretched length $l_1$.

The net force acting on the spring when the mass is attached to it and it is in equilibrium position is zero because the downward pull of gravity on the mass is counter equally by the spring force in opposite direction.

  $mg+\left[-k\left(l_1-l_0\right)\right]=0$        (I)

Here, m is the mass, g is the acceleration due to gravity, and k is the spring constant.

Now, when the spring is pulled further down by a distance x beyond its new equilibrium position, then net force on the mass is,

  $\begin{array}{c}{F}_{\text{net}}=-k\left(l_1-l_0\right)+mg+\left(-kx\right)\\ =0+\left(-kx\right)\\ =-kx\end{array}$

Substitute 0 for $-k\left(l_1-l_0\right)+mg$ in above equation.

  $\begin{array}{c}{F}_{\text{net}}=0+\left(-kx\right)\\ =-kx\end{array}$

Conclusion:

Therefore, the net force acting on the spring is $F_{\text{net}}=-kx$

(b)

To determine

To prove the net potential energy has the form $U\left(x\right)=\frac{1}{2}k{x}^2+\text{const}$.

Answer to Problem 5.1P

The net potential energy has the form $U\left(x\right)=\frac{1}{2}k{x}^2+\text{const}$.

Explanation of Solution

The stretching force is conservative as the net stretching force depends only on the distance the spring is stretched, that is x.

Write the expression for the potential energy.

    $U\left(x\right)=-{\displaystyle {\int }_0^x{F}_{\text{net}}\left(x\right)dx}$

Substitute $-kx$ for $F_{\text{net}}\left(x\right)$ in the above equation and simplify.

    $\begin{array}{c}U\left(x\right)=-{\displaystyle {\int }_0^x{F}_{\text{net}}\left(x\right)dx}\\ =-{\displaystyle {\int }_0^x\left(-kx\right)}dx\\ =k{\displaystyle {\int }_0^{x}x}dx\\ =k{\left(\frac{x^2}{2}\right)}_0^x+\text{Const}\\ =\frac{1}{2}k{x}^2+\text{Const}\end{array}$

Conclusion:

Therefore, the net potential energy has the form $U\left(x\right)=\frac{1}{2}k{x}^2+\text{const}$.