Question

A 10-kg mass hanging from a spring whose equilibrium length is 20 cm stretches it by 4.0 cm. A fish is then hung from the same spring and stretches if by 3.0 cm.

Find: A) What is the weight of the fish?

B) How long would the spring be stretched if the mass and the fish were hung together from the same spring?

Answer 1

mass of the object m=10kg

equilibrium length L=20cm

elongation x=4cm

now,

F=k*x

mg=k*x

k=(mg/x)

=10*9.8/4*10^-2

=2450 N/m

now,

let

mass of the fish is m'

weight of the fish is F'=m'*g

and elongation is x'=3cm

F'=k*x'

=2450*3*10^-2

=73.5 N

and

mass of the fish m'=F'/g

=73.5/9.8

=7.5 kg

B)

now,

F=k*x

(m+m')*g=k*x

(10+7.5)*9.8=2450*x

====>

x=0.07 m

=7 cm