Question

A fish (8 kg) is being yanked vertically upward out of the water and the fishing line breaks. If the line is rated to a maximum tension of 160 N (~ 35 lb test), then what was the minimum acceleration of the fish (in m/s2)

Answer 1

The minimum acceleration of the fish is 10.2 m/s²

Step-by-step explanation:

The force on string is:

F = 8 * 9.8 = 78.4 N

According the Newton second law:

T - W = ma

Where T = tension of the string = 160 N

W = weight of the fish = 78.4 N

m = mass of the fish = 8 kg

a = acceleration of the fish = ?

Clearing a:

a = (T- W)/m = (160 - 78.4)/8 = 10.2 m/s²

Answer 2

Step-by-step explanation:

Given:

Mass of the fish, 8 kg

Tension of the line, F = 160 N

Tension of the line, F + normal force = 0

Normal force, Fn = Mass × acceleration

Acceleration = force/mass

= 160/8

= 20 m/s^2

Minimum acceleration = 20 m/s^2