Question

Fish are hung on a spring scale to determine their mass (most fishermen feel no obligation to report the mass truthfully).

(a)

What is the force constant (in N/m) of the spring in such a scale if it stretches 8.30 cm for a 12.5 kg load?

N/m

(b)

What is the mass (in kg) of a fish that stretches the spring 5.50 cm?

kg

(c)

How far apart (in mm) are the half-kilogram marks on the scale?

mm

Answer 1

Part A.

Using Force balance on fish:

Force due to spring = Force due to gravity

Fs = Fg

k*x = m*g

k = m*g/x

Given that x = stretch in spring = 8.30 cm = 0.083 m

m = mass of spring = 12.5 kg

So,

k = 12.5*9.81/0.083

k = spring constant = 1477.41 N/m = 1.48*10^3 N/m

Part B.

for same spring when stretch is 5.50 cm

x = 0.055 m

then

m = k*x/g

m = 1477.41*0.055/9.81

m = 8.28 kg = mass of fish

Part C.

Half-kilogram marks on the scale will be apart by:

x = m*g/k

x = 0.5*9.81/1477.41

x = 3.32*10^-3 m = 3.32 mm

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