Question

a spring is stretched a distance of 0.05 meters from its equilibrium position. if the spring has a spring constant equal to 195 n/m, what must be the magnitude of the spring force exerted by this spring?

Explanation:

Step1: Identify the formula

Hooke's Law: $F = kx$, where $F$ is the spring force, $k$ is the spring constant, and $x$ is the displacement from equilibrium.

Step2: Substitute the values

Given $k = 195$ N/m and $x = 0.05$ m.
$F=(195)(0.05)$

Step3: Calculate the result

$F = 9.75$ N

Answer:

$9.75$ N