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 53.5 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 6.20 m/s after two-thirds of its length has left the surface, assume constant acceleration and determine the following.
(a) the salmon's acceleration?
(b) the magnitude of the force F during this interval?
god knows
a) 14.72
b) 787.52
To solve this problem, we will use the following equations of motion:
1. v = u + at
2. v^2 = u^2 + 2as
(a) To find the salmon's acceleration, we need to find the change in velocity and the time it takes to achieve this change.
1. The change in velocity is given by (final velocity - initial velocity):
Δv = 6.20 m/s - 3.00 m/s
= 3.20 m/s
2. We know that two-thirds of the fish's length has left the surface when it reaches a velocity of 6.20 m/s. Let's denote this length as 'd'.
3. The time taken to reach this velocity can be calculated using the equation:
v = u + at
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values:
t = (6.20 m/s - 3.00 m/s) / a
t = 3.20 m/s / a
4. We know that the distance 'd' is equal to two-thirds of the salmon's length, which is 1.50 m. Therefore, d = (2/3) * 1.50 m.
5. Using the equation of motion, v^2 = u^2 + 2as, we can solve for acceleration 'a'.
(6.20 m/s)^2 = (3.00 m/s)^2 + 2a * [(2/3) * 1.50 m]
Solving for 'a', we get:
a = [(6.20 m/s)^2 - (3.00 m/s)^2] / [2 * (2/3) * 1.50 m]
(b) To find the magnitude of the force F during this interval, we can use Newton's second law of motion:
F = m * a
Substituting the given values for mass and acceleration, we can calculate the force 'F'.
Now let's calculate the values step-by-step:
(a) Calculating the acceleration:
Δv = 3.20 m/s
d = (2/3) * 1.50 m = 1.00 m (rounded to two decimal places)
Using equation v^2 = u^2 + 2as to solve for 'a':
(6.20 m/s)^2 = (3.00 m/s)^2 + 2a * 1.00 m
a = [(6.20 m/s)^2 - (3.00 m/s)^2] / (2 * 1.00 m)
a ≈ 14.52 m/s^2 (rounded to two decimal places)
Therefore, the salmon's acceleration is approximately 14.52 m/s^2.
(b) Calculating the magnitude of the force F:
Given mass of the salmon, m = 53.5 kg
Using Newton's second law of motion, F = m * a:
F = 53.5 kg * 14.52 m/s^2
F ≈ 774.02 N (rounded to two decimal places)
Therefore, the magnitude of the force F during this interval is approximately 774.02 N.
To answer these questions, we need to use the principles of kinematics and the equations of motion. Let's break down the problem step by step:
(a) To find the salmon's acceleration, we can use the equation:
v^2 = u^2 + 2as
Where:
- v is the final velocity (6.20 m/s),
- u is the initial velocity (3.00 m/s),
- a is the acceleration (which we need to find),
- s is the distance traveled.
In this case, the distance traveled corresponds to two-thirds of the salmon's length, which is 2/3 * 1.50 m = 1.00 m.
Plugging in the given values into the equation, we have:
(6.20 m/s)^2 = (3.00 m/s)^2 + 2a * 1.00 m
Simplifying the equation further, we get:
38.44 m^2/s^2 = 9.00 m^2/s^2 + 2a
Subtracting 9.00 m^2/s^2 from both sides:
29.44 m^2/s^2 = 2a
Finally, we can solve for the acceleration a:
a = 29.44 m^2/s^2 / 2
a ≈ 14.72 m/s^2
Therefore, the salmon's acceleration is approximately 14.72 m/s^2.
(b) To determine the magnitude of the force F, we can use Newton's second law of motion:
F = m * a
Where:
- F is the force (which we need to find),
- m is the mass of the salmon (53.5 kg),
- a is the acceleration (which we found in part a, 14.72 m/s^2).
Plugging in the given values into the equation, we get:
F = 53.5 kg * 14.72 m/s^2
F ≈ 788.72 N
Therefore, the magnitude of the force F during this interval is approximately 788.72 Newtons.