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 $31.0 \mathrm {~cm} \text { ? }$

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, $ F = k \times 0.31 \, \text{m} $. Solving gives $ F = 199 \, \text{N} $.

An Ideal Spring Stretches by 21 31.0 cm ? 31.0 \mathrm {~cm} \text { ? }31.0 cm ?

An Ideal Spring Stretches by 21 $31.0 \mathrm {~cm} \text { ? }$