Fish are hung on a spring scale to determine their mass (most fishermen feel no obligation to truthfully report the mass).(a) What is the force constant of the spring in such a scale if it the spring stretches 8.00 cm fora 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?

(a) 1225 N/m

(b) 6.875 kg

(c) 0.4 cm

Explanation:

From hook's law,

F = ke Equation 1

Where F = Force, k = spring constant, e = extension.

From the question,

Note: The force acting on the spring is the weight of the fish.

F = mg.

Where m = mass of the fish, g = acceleration due to gravity.

mg = ke Equation 2

make k the subject of the equation,

k = mg/e Equation 3

Also make m the subject of the equation,

m = ke/g Equation 4

also make e the subject of the equation

e = mg/k Equation 5

(a)

using equation 3

k = mg/e

Given:

e = 8 cm = 0.08 cm, m = 10 kg, g = 9.8 m/s²

Substitute into equation 2

k = (10×9.8)/0.08

k = 1225 N/m.

(b)

If it stretches by 5.5 cm,

Given: e = 5.5 cm = 0.055 cm, k = 1225 N/m, g = 9.8 m/s²

Substitute into equation 4

m = (1225×0.055)/9.8

m = 6.875 kg.

(c)

Given: m = 0.5 kg, k = 1225 N/m, g = 9.8 m/s²

Substitute into equation 5

e = (0.5×9.8)/1225

e = 0.004 m

e = 0.4 cm.