A 42.0 kg swimmer with an initial speed of 1.50 m/s decides to coast until she comes to rest. If she slows with constant acceleration and stops after coasting 2.00 m, what was the magnitude of the force exerted on her by the water?
Well, well, well, it seems we have a swimmer who needs a physics lesson. Let's dive in, shall we?
To determine the magnitude of the force exerted on our water enthusiast, we can use Newton's second law, F = ma. Here, the swimmer's mass (m) is given as 42.0 kg.
The swimmer starts with an initial speed of 1.50 m/s and comes to rest after coasting for 2.00 m. This means her final velocity (v) would be 0 m/s.
Since the swimmer experiences constant deceleration, we can use the following kinematic equation: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
Plugging in the given values: 0^2 = 1.50^2 + 2a(2.00)
To find the acceleration (a), we can rearrange this equation and solve for a. Do you have a pen and paper ready? Just kidding, I can't see what you have, but here's the solution.
When we simplify, we get: 0 = 2.25 + 4a
Now, isolate 'a': a = -0.5625 m/s^2
Negative acceleration? Looks like the swimmer is slowing down. She must have swallowed too much water!
Finally, we can use Newton's second law, F = ma, to find the magnitude of the force. Substituting the mass (42.0 kg) and acceleration (-0.5625 m/s^2), we have:
F = (42.0 kg) * (-0.5625 m/s^2)
And after a little math, we get:
F = -23.62 N
Well, well, well, seems like the water had a grudge against our swimmer and exerted a force of approximately 23.62 N to bring her to a stop.
To find the magnitude of the force exerted on the swimmer by the water, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
Given:
Mass of the swimmer (m) = 42.0 kg
Initial speed of the swimmer (v₀) = 1.50 m/s
Final speed of the swimmer (v) = 0 m/s
Distance traveled by the swimmer (d) = 2.00 m
We need to find the acceleration of the swimmer first. We can use the formula:
v² = v₀² + 2ad
Rearranging the formula to solve for acceleration (a):
a = (v² - v₀²) / (2d)
Let's plug in the given values into the formula:
a = (0² - 1.50²) / (2 * 2.00)
a = (-2.25) / 4.00
a = -0.5625 m/s²
The negative sign indicates that the acceleration is in the opposite direction to the initial motion.
Now that we have the acceleration, we can use Newton's second law of motion to find the force:
F = m * a
Plugging in the values:
F = 42.0 kg * (-0.5625 m/s²)
F = -23.625 N
The magnitude of the force exerted on the swimmer by the water is 23.625 N.
To find the magnitude of the force exerted on the swimmer by the water, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
First, let's determine the acceleration of the swimmer. We can use the following kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity (0 m/s in this case), u is the initial velocity (1.50 m/s), a is the acceleration, and s is the displacement (2.00 m).
Rearranging the equation to solve for acceleration, we have:
a = (v^2 - u^2) / (2s)
Substituting the given values into the equation, we get:
a = (0^2 - 1.50^2) / (2 * 2.00)
a = (-1.50^2) / 4.00
a = -2.25 / 4.00
a = -0.5625 m/s^2
Note: The negative sign indicates that the swimmer is decelerating, or slowing down.
Now that we have the acceleration, we can calculate the force using Newton's second law:
F = m * a
Substituting the given mass of the swimmer (42.0 kg) and the calculated acceleration (-0.5625 m/s^2) into the equation, we have:
F = 42.0 kg * (-0.5625 m/s^2)
F = -23.625 N
The magnitude of the force exerted on the swimmer by the water is therefore 23.6 N.
F = m*a
a = (V^2-Vo^2)/2d
a = (0-(1.5^2))/4 = -0.5625 m/s^2.
F = 42 * (-0.5625) = -23.63 N.
The negative sign means the force opposes the motion.