Chapter 6, Problem 65P

(a)

To determine

The spring constant.

Answer to Problem 65P

Spring constant of the spring is $1.9\text{N}/\text{cm}$.

Explanation of Solution

Write the expression for the force acting on the spring.

$F=kx$

Here, $F$ is the force acting on the spring, $k$ is the spring constant and $x$ is the elongation or compression for the spring.

Write the expression for the weight hanging on the spring.

$W=mg$

Here, $W$ is the weight of the mass, $m$ is the mass that is hanging in equilibrium from the spring and $g$ is the acceleration due to gravity.

Since weight is hanging from the string. Force acting on the spring is equal to weight of the mass.

Therefore, write the expression for force acting on the spring.

$mg=kx$

Rearrange above equation to get $k$.

$k=\frac{mg}{x}$                                                                                                                          (I)

Conclusion:

Substitute $1.4\text{kg}$ for $m$, $9.80\text{N}/\text{kg}$ for $g$ and $7.2\text{cm}$ for $x$ in above equation to get $k$.

$\begin{array}{l}k=\frac{\left(1.4\text{kg}\right)\left(9.80\text{N}/\text{kg}\right)}{\left(7.2\text{cm}\right)}\\ =1.9\text{N}/\text{cm}\end{array}$

Therefore, spring constant of the spring is $1.9\text{N}/\text{cm}$.

(b)

To determine

The elastic potential energy stored in the spring.

Answer to Problem 65P

The elastic potential energy stored in the spring is $0.49\text{J}$.

Explanation of Solution

The elastic potential energy stored in the spring.

$U_{\text{elastic}}=\frac{1}{2}k{x}^2$

Here, $U_{\text{elastic}}$ is the elastic potential energy stored in the spring.

Substitute $\frac{mg}{x}$ for $k$ in above equation to get $U_{\text{elastic}}$.

$\begin{array}{c}{U}_{\text{elastic}}=\frac{1}{2}\left(\frac{mg}{x}\right)x^2\\ =\frac{1}{2}mgx\end{array}$

Conclusion:

Substitute $1.4\text{kg}$ for $m$ , $9.80\text{N}/\text{kg}$ for $g$ and $7.2\text{cm}$ for $x$ in above equation to get $U_{\text{elastic}\text{.}}$.
$\begin{array}{c}{U}_{\text{elastic}}=\frac{1}{2}\left(1.4\text{kg}\right)\left(9.80\text{N}/\text{kg}\right)\left(7.2\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =0.49\text{J}\end{array}$

Therefore, the elastic potential energy stored in the spring is $0.49\text{J}$.

(c)

To determine

The second mass if it can produce an elongation of $12.2\text{cm}$.

Answer to Problem 65P

The second mass if it can produce an elongation of $12.2\text{cm}$ is $2.4\text{kg}$.

Explanation of Solution

Let $m^{\prime }$ be the new mass that hanging from spring and $x^{\prime }$ is the elongation due to new mass $m^{\prime }$.

Write the expression for the spring constant in terms of new mass and elongation.

$k=\frac{m^{\prime }g}{x^{\prime }}$ (II)

Spring constant of the spring is a constant independent of force.

Conclusion:

Divide equation (II) by (I) to get $m^{\prime }$.

$\begin{array}{l}\frac{k}{k}=\frac{\frac{m^{\prime }g}{x^{\prime }}}{\frac{mg}{x}}\\ m^{\prime }=\frac{m{x}^{\prime }}{x}\end{array}$ (III)

Substitute $1.4\text{kg}$ for $m$ , $12.2\text{cm}$ for $x_2$ and $7.2\text{cm}$ for $x_1$ in above equation to get $m^{\prime }$.

$\begin{array}{c}{m}^{\prime }=\frac{\left(1.4\text{kg}\right)\left(12.2\text{cm}\right)}{\left(7.2\text{cm}\right)}\\ =2.4\text{kg}\end{array}$

Therefore, the second mass if it can produce an elongation of $12.2\text{cm}$ is $2.4\text{kg}$.