Question

A fish scale, consisting of a spring with spring constant k=200N/m, is hung vertically from the ceiling. A 4.9 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. Part A What is the maximum distance through which the fish falls?

Answer 1

The maximum distance through which the fish falls is calculated using energy conservation, equating the gravitational potential energy to the elastic potential energy of the spring, resulting in a maximum fall distance of 69.3 cm.

Step-by-step explanation:

To find the maximum distance through which the 4.9 kg fish falls on the hanging spring, we'll use the concept of energy conservation. Since the fish is released from rest, all the potential energy (due to its height) will be converted into elastic potential energy at the maximum stretch of the spring.

The potential energy (PE) of the fish when it is at the equilibrium position (unstretched spring) just before being released is given by PE = m * g * h, where m is the mass of the fish, g is the acceleration due to gravity (9.8 m/s²), and h is the height which we are trying to find.

The elastic potential energy (EPE) of the spring at its maximum stretch is given by EPE = 1/2 * k * x², where k is the spring constant and x is the maximum stretch of the spring.

Setting these equal to each other gives us m * g * h = 1/2 * k * x². Now we solve for x to find the maximum stretch:

• m = 4.9 kg
• g = 9.8 m/s²
• k = 200 N/m

h = (1/2 * k * x²) / (m * g)

Rearranging and solving for x we get:

x = √(2 * m * g / k)

x = √(2 * 4.9 kg * 9.8 m/s² / 200 N/m)

x = √(96.04 / 200)

x = √0.4802

x = 0.693 m or 69.3 cm

Thus, the maximum distance the fish falls is 69.3 cm.