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Ch 05: Applying Newton's Laws
Chapter 5, Problem 5

You throw a baseball straight upward. The drag force is proportional to υ2. In terms of g, what is the y-component of the ball's acceleration when the ball's speed is half its terminal speed and (a) it is moving up?

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Identify the forces acting on the baseball: When the baseball is thrown upward, two main forces act on it - the gravitational force (weight) downward and the drag force upward. The drag force is given as proportional to the square of the velocity (υ^2).
Set up the equation for net force: The net force acting on the baseball can be expressed as F_net = mg - kv^2, where m is the mass of the baseball, g is the acceleration due to gravity, k is a constant of proportionality, and v is the velocity of the baseball.
Determine the terminal velocity: Terminal velocity occurs when the net force on the object is zero (F_net = 0). This gives mg = kv_t^2, where v_t is the terminal velocity. Solve for v_t to find v_t = \sqrt{\frac{mg}{k}}.
Calculate the velocity when it is half the terminal speed: When the speed is half the terminal speed, v = \frac{1}{2}v_t. Substitute v_t from the previous step to find v = \frac{1}{2}\sqrt{\frac{mg}{k}}.
Find the acceleration: Substitute v = \frac{1}{2}v_t into the net force equation and solve for the acceleration a. The net force equation becomes F_net = m(g - k(\frac{1}{2}v_t)^2). Simplify and solve for a to find the y-component of the ball's acceleration when moving up.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Terminal Velocity

Terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium prevents further acceleration. For a baseball thrown upward, terminal velocity occurs when the gravitational force is balanced by the drag force, resulting in zero net acceleration.
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Drag Force

The drag force is a resistive force acting opposite to the direction of motion of an object moving through a fluid, such as air. In this scenario, the drag force is proportional to the square of the velocity (υ²), meaning it increases significantly as the speed of the baseball increases, affecting its acceleration.
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Acceleration due to Gravity

Acceleration due to gravity (g) is the acceleration experienced by an object due to the gravitational pull of the Earth, approximately 9.81 m/s² downward. When analyzing the motion of the baseball, this acceleration must be considered alongside the drag force to determine the net acceleration of the ball at different speeds.
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Related Practice
Textbook Question
The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s).(c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero?
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Textbook Question
The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s). (d) What then would be the passenger's apparent weight at the lowest point?
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Textbook Question
flat (unbanked) curve on a highway has a radius of 170.0 m. A car rounds the curve at a speed of 25.0 m/s. (a) What is the minimum coefficient of static friction that will prevent sliding?
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Textbook Question
You throw a baseball straight upward. The drag force is proportional to υ2. In terms of g, what is the y-component of the ball's acceleration when the ball's speed is half its terminal speed and (b) It is moving back down?
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Textbook Question
A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 N. Then you must reduce your push to 208 N to keep it moving at a steady 25.0 cm/s. (a) What are the coefficients of static and kinetic friction between the crate and the floor?
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Textbook Question
A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 N. Then you must reduce your push to 208 N to keep it moving at a steady 25.0 cm/s. (b) What push must you exert to give it an acceleration of 1.10 m/s2?
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