A fish (6 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 150 N (~ 33 lb test), then what was the minimum acceleration of the fish (in m/s2)?

15.2 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S.I unit of acceleration is m/s²

From the question,

F-W = ma Equation 1

Where F = Tension on the line, W = weight of the fish, m = mass of the fish, a = acceleration of the fish.

But,

W = mg Equation 2

Where g = acceleration due to gravity.

Substitute equation 2 into equation 1

F-mg = ma

make a the subject of the equation

a = (F-mg)/m Equation 3

Given: F = 150 N, m = 6 kg,

Constant: g = 9.8 m/s²

Substitute into equation 2

a = [150-(6×9.8)]/6

a = (150-58.8)/6

a = 91.2/6

a = 15.2 m/s²

Hence the minimum acceleration of the fish = 15.2 m/s²

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s