Solve the problem.
-A bathroom scale is compressed $\frac { 1 } { 4 }$ in. when a $220 \mathrm { lb }$ person stands on it. Assuming that the scale behaves like a spring that obeys Hooke's law, how much does someone who compresses the scale $\frac { 1 } { 8 }$ in. weigh?
A) 110 lb
B) 55 lb
C) 440 lb
D) 165 lb

Question

Quizplus · Accepted Answer

Hooke's law states that the force $ F $ needed to extend or compress a spring by some distance $ x $ is proportional to that distance. That is, $ F = kx $, where $ k $ is a constant specific to the spring. In this case, when the scale is compressed $\frac{1}{4}$ inch, it supports a weight of 220 lb. This means $ 220 = k \times \frac{1}{4} $. To find the weight of a person who compresses the scale $\frac{1}{8}$ inch, we first find $ k $ and then use it to find the new weight. Since $ k = \frac{220}{\frac{1}{4}} = 880 $, the weight for a $\frac{1}{8}$ inch compression is $ F = 880 \times \frac{1}{8} = 110 $ lb.

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