A force of 50 N is required to hold a spring that has been stretched from its natural length of 10 cm to 60 cm. How much work (in N . m) had to be done to stretch it that far?
A) 12.5 N . m
B) 14.0 N . m
C) 25.0 N . m
D) 6.25 N . m
E) 14.6 N . m

Question

Quizplus · Accepted Answer

The work done to stretch the spring can be calculated using the formula for the work done in stretching (or compressing) a spring: $W = \frac{1}{2} k x^2$, where $W$ is the work done, $k$ is the spring constant, and $x$ is the displacement from the natural length. Given that a force of 50 N is required to hold the spring at a stretch of 50 cm (0.5 m from its natural length), we can first find the spring constant using Hooke's Law ($F = kx$), which gives $k = \frac{F}{x} = \frac{50}{0.5} = 100 \, \text{N/m}$. Substituting $k = 100 \, \text{N/m}$ and $x = 0.5 \, \text{m}$ into the work formula gives $W = \frac{1}{2} \times 100 \times (0.5)^2 = 12.5 \, \text{N} \cdot \text{m}$.

A Force of 50 N Is Required to Hold a Spring