Chapter 12, Problem 1PLA

To determine

The force exerted by the spring.

Answer to Problem 1PLA

The force exerted by the spring is $0.413\text{ N}$.

Explanation of Solution

Write the expression for force using Hooke’s law

    $F=-kx$        (I)

Here, $F$ is the force, $k$ is the spring constant and $x$ is the displacement of spring.

Substitute $7.50\text{N}/\text{m}$ for $k$ and $0.550\text{ m}$ for $x$ in equation (I) to find the value of $F$

$\begin{array}{c}F=-\left(7.50\text{N}/\text{m}\right)\left(0.550\text{ m}\right)\\ =-0.413\text{ N}\end{array}$

The force exerted by the spring is $0.413\text{ N}$ and the negative sign indicates that it is a restoring force of the spring.

Conclusion:

Thus, the force exerted by the spring is $0.413\text{ N}$.