Chapter 6, Problem 126P

(a)

To determine

The effective spring constant of the combination.

Answer to Problem 126P

Theeffective spring constant of the combination is $\frac{1}{1/k_1+1/k_2}$.

Explanation of Solution

Imagine that spring (1) is suspended from a ceiling and spring (2) is attached to the bottom of the first spring. An object of mass $m$ is attached to the bottom end of the second spring.

Write the Hooke’s law for the spring (1).

$F_1=k_1{x}_1$ (I)

Here, $F_1$ is the spring force of spring (1), $k_1$ is its spring constant and $x_1$ is the stretching of spring (1)

Rewrite the above equation for $x_1$ .

$x_1=\frac{F_1}{k_1}$ (II)

Write the Hooke’s law for the spring (2).

$F_2=k_2{x}_2$ (III)

Here, $F_2$ is the spring force of spring (2), $k_2$ is its spring constant and $x_2$ is the stretching of spring (2)

Rewrite the above equation for $x_2$ .

$x_2=\frac{F_2}{k_2}$ (IV)

Assume that the masses of the springs are negligible and the system is in equilibrium. The mass connected to the spring (2) exerts a force on the lower spring equal to its weight.

$F_2=W$ (V)

Here, $W$ is the weight of the object attached at the lower spring

Put the above equation in equation (IV).

$x_2=\frac{W}{k_2}$ (VI)

Write the equation for the force the lower spring exerts on spring(1) .

$F_1=W$

Put the above equation in equation (II).

$x_1=\frac{W}{k_1}$

Put equation (V) in the above equation.

$x_1=\frac{F_2}{k_1}$ (VII)

Compare equations (II) and (VII).

$F_1=F_2$

Put equations (I) and (III) in the above equation.

$k_1{x}_1=k_2{x}_2$ (VIII)

Take $F_1=F_2=F$ and $x=x_1+x_2$ . Imagine the two springs in series as only one spring which stretches an amount $x$ in response to a force $F$ .

$x=x_1+x_2$

Put equations (II) and (IV) in the above equation.

$\begin{array}{c}x=\frac{F_1}{k_1}+\frac{F_2}{k_2}\\ =\frac{F}{k_1}+\frac{F}{k_2}\\ =F\left(\frac{1}{k_1}+\frac{1}{k_2}\right)\end{array}$

Rewrite the above equation for $F$ .

$\begin{array}{c}F=x{\left(\frac{1}{k_1}+\frac{1}{k_2}\right)}^{-1}\\ =\frac{1}{1/k_1+1/k_2}x\\ F=kx\\ \Rightarrow k=\frac{1}{1/k_1+1/k_2}\end{array}$

Conclusion:

Therefore, the effective spring constant of the combination is $\frac{1}{1/k_1+1/k_2}$ .

(b)

To determine

The potential energy stored in the spring.

Answer to Problem 126P

Thepotential energy stored in the springis $0.15\text{ J}$ .

Explanation of Solution

Write the equation for the potential energy stored.

$U=\frac{1}{2}k{x}^2$

Here, $U$ is the potential energy stored in the spring

In part (a), it is found that the expression for $k$ is $\frac{1}{1/k_1+1/k_2}$.

Substitute $\frac{1}{1/k_1+1/k_2}$ for $k$ in the above equation.

$U=\frac{1}{2}\frac{1}{1/k_1+1/k_2}{x}^2$ (IX)

Conclusion:

Given that the value of $k_1$ is $5.0\text{ N/cm}$ , the value of $k_2$ is $3.0\text{ N/cm}$ and the value of $x$ is $4.0\text{ cm}$ .

Substitute $5.0\text{ N/cm}$ for $k_1$ , $3.0\text{ N/cm}$ for $k_2$ and $4.0\text{ cm}$ for $x$ in equation (IX) to find $U$ .

$\begin{array}{c}U=\frac{1}{2}\frac{1}{1/\left(5.0\text{ N/cm}\right)+1/\left(3.0\text{ N/cm}\right)}{\left(4.0\text{ cm}\right)}^2\\ =0.15\text{ J}\end{array}$

Therefore, the potential energy stored in the spring is $0.15\text{ J}$ .