An ideal spring stretches by 21.0 cm when a 135-N object is hung from it.If instead you hang a fish from this spring,what is the weight of a fish that would stretch the spring by $31.0 \mathrm {~cm} ?$
A)199 N
B)91 N
C)145 N
D)279 N

Question

Quizplus · Accepted Answer

The spring's stretch is directly proportional to the force applied to it, following Hooke's Law. Given that a 135-N object stretches the spring by 21.0 cm, a stretch of 31.0 cm would be achieved by a force (weight in this context) that is in the same ratio, i.e., $135 \, \text{N} \times \frac{31.0 \, \text{cm}}{21.0 \, \text{cm}} = 199 \, \text{N}$.

An Ideal Spring Stretches by 21 31.0 cm?31.0 \mathrm {~cm} ?31.0 cm? A)199 N B)91 N C)145 N D)279 N

An Ideal Spring Stretches by 21 $31.0 \mathrm {~cm} ?$ A)199 N
B)91 N
C)145 N
D)279 N