Chapter 1.11, Problem 29E

Hooke’s Law Hooke’s Law states that the force needed to keep a spring stretched x units beyond its natural length is directly proportional to x. Here the constant of proportionality is called the spring constant.

1. (a) Write Hooke’s Law as an equation.
2. (b) If a spring has a natural length of 5 cm and a force of 30 N is required to maintain the spring stretched to a length of 9 cm, find the spring constant.
3. (c) What force is needed to keep the spring stretched to a length of 11 cm?

To determine

To express: Hooke’s law for spring as equation.

Answer to Problem 29E

Hooke’s law for spring is $F=kx$.

Explanation of Solution

Hooke’s law:

The force required to keep the spring stretched beyond the natural length is directly proportional to the extension of spring beyond natural length.

$F$ is directly proportional to $x$

Mathematically,

$F=kx$

Here,

$F$ is force needed to keep the spring stretched.

$k$ is spring constant.

$x$ is the extension of spring beyond natural length.

Thus, Hooke’s law as an equation is $F=kx$.

To determine

To find: The spring constant $k$.

Answer to Problem 29E

The value of spring constant is $7.5\frac{\text{N}}{\text{cm}}$.

Explanation of Solution

Given:

Natural length of spring ( $x_0$) is $5\text{cm}$.

Force required to stretch it up to $9\text{cm}$ is $30\text{N}$.

Calculation:

The spring is stretched up to $9\text{cm}$, as the natural length is $5\text{cm}$, it is stretched $4\text{cm}$ beyond the natural length. So, the value of $x$ is $4\text{cm}$.

Use Hooke’s Law,

$F=kx$

Substitute, 30 for $F$ and 4 for $x$ in Hooke’s law equation and solve for $k$.

$30=k\cdot 4$

Multiply both the sides by $\left(\frac{1}{4}\right)$

$\begin{array}{c}\left(\frac{1}{4}\right)\cdot 30=\left(\frac{1}{4}\right)k\cdot 4\\ 7.5=k\end{array}$ (1)

Thus, the value of $k$ is $7.5\frac{\text{N}}{\text{cm}}$.

To determine

To find: The force required to keep the spring stretched to $11\text{cm}$.

Answer to Problem 29E

The force required is $45\text{N}$.

Explanation of Solution

Given:

The stretched length of the spring is $11\text{cm}$.

Calculation:

The natural length of spring is $5\text{cm}$, but spring is stretched to a length of $11\text{cm}$ i.e. $6\text{cm}$ beyond the natural length. So, the value of $x$ is $6\text{cm}$.

Use Hooke’s Law,

$F=kx$

Substitute 6 for $x$ and 7.5 for $k$ from equation (1) and solve for $F$.

$\begin{array}{c}F=\left(7.5\right)\cdot \left(6\right)\\ F=45\end{array}$

Thus, the value of $F$ is $45\text{N}$.