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 constant $ k $ can be calculated using Hooke's Law $ F = kx $. For the 135-N object, $ k = \frac{135 \, \text{N}}{0.21 \, \text{m}} $. For the fish, using the same $ k $, the force $ F $ is $ F = k \times 0.31 \, \text{m} $, which calculates to approximately 199 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)

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