A chinook salmon has a maximun underwater speed of 3.0 m/s, and can jump out of

Question

A chinook salmon has a maximun underwater speed of 3.0 m/s, and can jump out of the water vertically with a speed of 6.0 m/s. A record salmon has a length of 1.5 m and a mass of 61 kg. When swimming updard at constant speed, and neglecting buoyancy, the fish exeriences three forces: an upward force F exerted by the tail fin, the downward drag force of the water, and the downward force of gravity. As the fish leaves the surface of the water, however, it experiences a net upward force causing it to accelerate from 3.0 m/s to 6.0 m/s. Assuming the drag force disappears as soon as the head of the fish breaks the surface and F is exerted until two-thirds of the fish's length has left the water, determine the magnitude of F.

Explanation / Answer

Calcboy 1357 explained this well GIVEN: salmon is 1.5 m long, 2/3 of its length is 1 m. This is the distance traveled during the acceleration from 3 m/s to 6.9m/s. EQUATIONS: The kinetic energy E of a body of mass m moving at velocity vis E = 1/2 * m * v^2 So the work W necessary to make the difference in kinetic energy of a salmon at 3 m/s and 6.9 m/s is W = 1/2 * 64 kg * (6.9 m/s)^2 - 1/2 * 69 kg * (3 m/s)^2= 1235.52 J Now work is force times distance W = F * d which you can solve for F F = W / d and inserting the distance (2/3 of the salmons length, seeabove): F = 1235.52 J / 1 m = 1235.52 N Nowthis is the net force necessary for the acceleration. This netforce is the sum of the tail fin force Ff and gravity Fg. F = Ff + Fg (the force of gravity acts downwards, so it will assume a negativevalue) Solving this for the tail fin force gives Ff = F - Fg Now gravity on a salmon of mass m = 61 kg is Fg = - g * m = - 9.81 m/s^2 * 64 kg = - 627.2 N So the tail fin force is

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