Question

A fish scale, consisting of a spring with spring constant k=200N/m, is hung vertically from the ceiling. A 2.6 kg fish is attached to the end of the unstretched spring and then released. The fish moves downward until the spring is fully stretched, then starts to move back up as the spring begins to contract.
What is the maximum distance through which the fish falls?

Answer 1

Newton's second law

• F=ma

Find weight of fish

• F=2.6(10)=26N

Now

$\\ \rm\Rrightarrow F=-kx$

$\\ \rm\Rrightarrow k=\dfrac{-F}{x}$

$\\ \rm\Rrightarrow x=\dfrac{-F}{k}[)tex]Keep it positive[tex]\\ \rm\Rrightarrow x=\dfrac{26}{200}$

$\\ \rm\Rrightarrow x=0.13m$

Answer 2

Answer:

Explanation:

The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.

Change in gravity potential energy = change in spring potential energy

mgh = 1/2kh^2

Assume gravity constant g is 10m/s^2

2.6*10*h = 1/2*200*h^2

100h^2 - 26h = 0

2h(50h - 13) = 0

h = 0 or h = 13/50 = 0.65m

h = 0 is before the spring is stretched

So the maximum distance is 0.65m.