Answer:
199.28 N
Explanation:
It is given that,
A spring stretches by 21.0 cm when a 135 N object is attached.
We need to find the weight of a fish that would stretch the spring by 31 cm.
We know that, the force in a spring is given by as per Hooke's law as follows :
F = kx
Where k is spring constant
[tex]\dfrac{F_1}{F_2}=\dfrac{x_1}{x_2}\\\\\dfrac{135}{F_2}=\dfrac{21}{31}\\\\F_2=\dfrac{135\times 31}{21}\\\\F_2=199.28\ N[/tex]
So, the required weight of a fish is 199.28 N.
