A Chinook salmon can swim underwater at 3.75 m/s, and it can also jump vertically upward, leaving the water with a speed of 6.20 m/s. A record salmon has length 1.50 m and mass 61.0 kg. Consider the fish swimming straight upward in the water below the surface of a lake. The gravitational force exerted on it is very nearly canceled out by a buoyant force exerted by the water. The fish experiences an upward force P exerted by the water on its threshing tail fin and a downward fluid friction force that we model as acting on its front end. Assume the fluid friction force disappears as soon as the fish's head breaks the water surface and assume the force on its tail is constant. Model the gravitational force as suddenly switching full on when half the length of the fish is out of the water. Find the value of P.

Question

Houston Rosales

Physics

14 August, 00:18

A Chinook salmon can swim underwater at 3.75 m/s, and it can also jump vertically upward, leaving the water with a speed of 6.20 m/s. A record salmon has length 1.50 m and mass 61.0 kg. Consider the fish swimming straight upward in the water below the surface of a lake. The gravitational force exerted on it is very nearly canceled out by a buoyant force exerted by the water. The fish experiences an upward force P exerted by the water on its threshing tail fin and a downward fluid friction force that we model as acting on its front end. Assume the fluid friction force disappears as soon as the fish's head breaks the water surface and assume the force on its tail is constant. Model the gravitational force as suddenly switching full on when half the length of the fish is out of the water. Find the value of P.

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Answers (1)

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Alexander Garcia · Answer 1

P = 2161 N

Explanation:

For this exercise, let's start with Newton's second law

P - W = m a

P = ma + W

Where P is the fin force, W the weight,

Let's look for the vertical acceleration of the fish, this goes from a vertical speed of zero to a speed of 6.20 m / s when it has traveled half its length 1.50

y = 0.75 m

v² = v₀ + 2 a y

a = v² / 2 y

a = 6.20²/2 0.75

a = 25.63 m / s

We substitute in Newton's second law

P = m (g + a)

Let's calculate

P = 61.0 (0 9.8 + 25.63)

P = 2161 N