Question

A scale used to weigh fish consists of a spring hung from a support. The spring's equilibrium length is 10.0 cm. When a 4.0 kg fish is suspended from the end of the spring, it stretches to a length of 13.4 cm.
(Part A) What is the spring constant k for this spring? Express your answer with the appropriate units.
(Part B) If an 8.0 kg fish is suspended from the spring, what will be the length of the spring? Express your answer with the appropriate units.

Answer 1

A) 1153 N/m

We can find the spring constant by using Hooke's law:

$F=kx$

where

F is the force applied to the spring

k is the spring constant

x is the displacement of the spring

In this problem, a fish of mass m = 4.0 kg is hanging on the spring, so the force applied is the weight of the fish:

$F=mg=(4.0 kg)(9.8 m/s^2)=39.2 N$

and the displacement of the spring is:

$x = 13.4 cm - 10.0 cm = 3.4 cm = 0.034 m$

so, the spring constant is

$k=\frac{F}{x}=\frac{39.2 N}{0.034 m}=1153 N/m$

B) 16.8 cm

In this case, a fish of mass

m = 8.0 kg

is hanging on the spring. Therefore, the force applied to the spring is

$F=mg=(8.0 kg)(9.8 m/s^2)=78.4 N$

So we can find the displacement of the spring:

$x=\frac{F}{k}=\frac{78.4 N}{1153 N/m}=0.068 m = 6.8 cm$

And since the equilibrium length of the spring is

$x_0 = 10.0 cm$

the new length of the spring will be

$x' = 10.0 cm + 6.8 cm = 16.8 cm$