2. As a fish jumps vertically out of the water, assume that only two significant
forces act on it: an upward force F exerted by the tail fin and the downward
force due to gravity. A record Chinook salmon has a length of 1.50 m and a
mass of 61.0 kg. If this fish is moving upward at 3.00 m/s as its head first
breaks the surface and has an upward speed of 6.00 m/s after two- thirds of
its length has left the surface, assume constant acceleration and determine
(a) the salmon’s acceleration and (b) the magnitude of the force F during this
interval.
To find the salmon's acceleration and the magnitude of the force during this interval, we can use the equations of motion.
(a) To find the salmon's acceleration, we can use the formula:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
In this case, the initial velocity (u) is 3.00 m/s and the final velocity (v) is 6.00 m/s. The displacement (s) is two-thirds of the salmon's length, which is (2/3) * 1.50 m = 1.00 m.
Rearranging the equation and plugging in the values, we have:
a = (v^2 - u^2) / (2s)
a = (6.00^2 - 3.00^2) / (2 * 1.00)
a = (36.00 - 9.00) / 2.00
a = 27.00 / 2.00
a = 13.50 m/s^2
Therefore, the salmon's acceleration during this interval is 13.50 m/s^2.
(b) To find the magnitude of the force, we can use Newton's second law of motion:
F = ma
where:
F = Force
m = mass
a = acceleration
In this case, the mass (m) of the salmon is given as 61.0 kg, and the acceleration (a) is 13.50 m/s^2.
Plugging in these values into the equation, we have:
F = 61.0 kg * 13.50 m/s^2
F = 823.50 N
Therefore, the magnitude of the force exerted by the tail fin (F) during this interval is 823.50 N.