An ideal spring with a spring constant (force constant)of $22 \mathrm {~N} / \mathrm { m }$ is stretched from equilibrium to 2.9 m. How much work is done in the process?
A)93 J
B)186 J
C)47 J
D)121 J

Question

Quizplus · Accepted Answer

The work done on a spring is given by the formula $ W = \frac{1}{2} k x^2 $, where $ k $ is the spring constant and $ x $ is the displacement. Substituting $ k = 22 \, \text{N/m} $ and $ x = 2.9 \, \text{m} $, we get $ W = \frac{1}{2} \times 22 \times (2.9)^2 = 92.57 \, \text{J} $, which rounds to 93 J.

An Ideal Spring with a Spring Constant (Force Constant)of 22 N/m22 \mathrm {~N} / \mathrm { m }22 N/m

An Ideal Spring with a Spring Constant (Force Constant)of $22 \mathrm {~N} / \mathrm { m }$