Question

a fisherman uses a spring scale to weigh a tilapia fish. he records the fish weight as a kilograms and notices that the spring stretches b centimeters. which expression represents the spring constant (1 kg = 9.8 n)?
\\(\frac{9.8a}{b}\\)
980ab
9.8ab
\\(\frac{980a}{b}\\)

Explanation:

Step1: Calculate the weight force

The weight of the fish in Newtons is $F = 9.8a$ (since $1\ kg=9.8\ N$ and the mass of the fish is $a$ kg). The stretch of the spring is $x = b$ cm. We need to convert the stretch to meters, so $x=\frac{b}{100}$ m.

Step2: Apply Hooke's Law

Hooke's Law is $F = kx$, where $F$ is the force applied, $k$ is the spring - constant, and $x$ is the displacement from the equilibrium position. Rearranging for $k$, we get $k=\frac{F}{x}$. Substituting $F = 9.8a$ N and $x=\frac{b}{100}$ m into the formula for $k$, we have $k=\frac{9.8a}{\frac{b}{100}}=\frac{980a}{b}$.

Answer:

$\frac{980a}{b}$