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?
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
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