Question

"a very light ideal spring stretches by 21.0 cm when it is used to hang a 135-n object. what is the weight of a piece of electronic equipment that would stretch the spring 44.9 cm by if you hung the equipment using the spring?

Answer 1

In the first scenario, upon reaching equilibrium
       mg = kx
      135 = k(0.21) [Converted x in SI units]
          k = 135/(0.21)

Similarly, in the second scenario, upon reaching equilibrium
       mg = kx
       mg = [(135)(.449)]/(0.21)
       mg = weight of the electronic equipment = 289 N(approx)

Answer 2

Answer:

F= 288.7 N

Explanation:

Hooks Law

"states that the strain/deformation of an elastic object or material is proportional to the stress applied to it."

From Hooke's Law we can determine the spring constant  

F = k*x  

so

k = F/x  

k=  135N/.21m

k= 643 N/m

So  if a fish stretches the spring 0.449m  then it would weigh  

F =k*x

F= 643N/m*0.449m

F= 288.707 N