Question

A scale used to weigh fish consists of a spring hung from a support. The spring's equilibrium length is 10.0 cm. When a 4.0 kg fish is suspended from the end of the spring, it stretches to a length of 13.2 cm.

1. What is the spring constant k for this spring? Express your answer with the appropriate units.

2. If an 8.0 kg fish is suspended from the spring, what will be the length of the spring? Express your answer with the appropriate units.

Answer 1

1) Lo = 10.0 cm
L = 13.2 cm
extension of the spring, x = L - Lo

= 13.2 - 10.0

= 3.2 cm

= 0.032 m

using eqyulibrium condition,
F_spring = Fg

k*x = m*g

k = m*g/x

= 4*9.8/0.032

= 1225 N/m <<<<<<<<<<---------------Answer

B) let x is the extension of the spring when 8 kg fish is suspended.

F_spring = Fg

k*x = m*g

x = m*g/k

= 8*9.8/1225

= 0.064 m

= 6.4 cm

New length, l = Lo + x

= 10.0 + 6.4

= 16.4 cm <<<<<<<<<<---------------Answer