Question

Fish are hung on a spring scale to determine their mass. (a) What is the force constant of the spring in such a scale if the spring stretches 8.00 cm for a 10.0 kg load? (b) What is the mass of a fish that stretches the spring 5.50 cm? (c) How far apart are the half-kilogram marks on the scale?

Answer 1

a) The spring constant is 1225 N/m

b) The mass of the fish is 6.88 kg

c) The marks are 0.4 cm apart

Explanation:

a)

When the spring is at equilibrium, the weight of the load applied to the spring is equal (in magnitude) to the restoring force of the spring, so we can write

$mg = kx$

where

m is the mass of the load

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x is the stretching of the spring

For the load in this problem we have

m = 10.0 kg

x = 8.00 cm = 0.08 m

Substituting, we find the spring constant

$k=\frac{mg}{x}=\frac{(10)(9.8)}{0.08}=1225 N/m$

b)

As before, at equilibrium, the weight of the fish must balance the restoring force in the spring, so we have

$mg=kx$

where this time we have:

m = mass of the fish

$g=9.8 m/s^2$

k = 1225 N/m is the spring constant

x = 5.50 cm = 0.055 m is the stretching of the spring

Substituting,

$m=\frac{kx}{g}=\frac{(1225)(0.055)}{9.8}=6.88 kg$

c)

To solve this part, we just need to find the change in stretching of the spring when a load of half-kilogram is hanging on the spring. Using again the same equation,

$mg=kx$

where this time we have:

m = 0.5 kg

$g=9.8 m/s^2$

k = 1225 N/m

x = ? is the distance between the half-kilogram marks on the scale

Substituting,

$x=\frac{mg}{k}=\frac{(0.5)(9.8)}{1225}=0.004 m = 0.4 cm$

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