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You throw a baseball straight up. The drag force is proportional to \(v^{2} .\) In terms of \(g,\) what is the \(y\) -component of the ball's acceleration when its speed is half its terminal speed and (a) it is moving up? It is moving back down?

Short Answer

Expert verified
(a) Moving up: Acceleration is \( \frac{5g}{4} \). Moving down: Acceleration is \( \frac{3g}{4} \).

Step by step solution

01

Understand the forces involved

The forces acting on the baseball are gravity, which acts downward with acceleration due to gravity \( g \), and the drag force, which opposes the direction of motion and is proportional to the square of velocity, \( F_d = kv^2 \). Terminal velocity is reached when the gravitational force equals the drag force.
02

Write equations for forces and accelerations

The terminal speed \( v_t \) is when gravitational force equals the drag force, \( mg = kv_t^2 \). Hence, \( k = \frac{mg}{v_t^2} \). Thus, for a speed \( v = \frac{v_t}{2} \), the drag force is \( F_d = k \left( \frac{v_t}{2} \right)^2 = \frac{mg}{v_t^2} \cdot \frac{v_t^2}{4} = \frac{mg}{4} \).
03

Find acceleration when moving up

When moving upwards, both gravity and drag force act downward. The net force is thus \( F_{ ext{net}} = mg + \frac{mg}{4} = \frac{5mg}{4} \). Using Newton's second law \( F = ma \), we find acceleration: \( a = \frac{F_{ ext{net}}}{m} = \frac{5g}{4} \).
04

Find acceleration when moving down

When moving downwards, gravity acts downward while drag force acts upwards. The net force is \( F_{ ext{net}} = mg - \frac{mg}{4} = \frac{3mg}{4} \). Using Newton's second law \( F = ma \), we find acceleration: \( a = \frac{F_{ ext{net}}}{m} = \frac{3g}{4} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Drag Force
When we talk about motion and forces, drag force is a common topic, especially in fluids like air or water. Drag force is the resistance a moving object faces in a fluid. Imagine throwing a ball straight up into the air. As it rises or falls, air pushes against it, slowing it down, and that's drag in action.
The drag force can be complicated because it often depends on the speed of the object. In this baseball example, the drag force is proportional to the square of the velocity, which means as the speed doubles, the drag force increases four times.
  • Formula: If the drag force is proportional to speed squared, it's expressed as: \( F_d = kv^2 \)
  • Where \( k \) is a constant that depends on the properties of the fluid and the shape of the object.
Understanding how drag works is crucial for calculating other forces, as it directly impacts the object's motion. In the upward motion of the baseball, both gravity and drag force act in the same direction (downward), increasing the total resistance it faces.
Gravitational Force
Gravitational force is a key player in motion problems like these. It's the force that attracts two bodies toward each other. On Earth, it's what keeps us grounded. For our baseball, gravity pulls it downward.
  • Formula: The force due to gravity is given by: \( F_g = mg \)
  • Where \( m \) is the mass of the baseball and \( g \) is the acceleration due to gravity, typically \( 9.8 \, m/s^2 \).
When you throw the baseball into the air, gravity tries to bring it back down. The gravitational force remains constant at all times, pulling with the same strength both up and down.
In the step-by-step solution, gravity combines with drag to determine the net force when the ball is moving in either direction.
Terminal Velocity
Terminal velocity is a concept that's all about balance. It's the constant speed that a freely falling object eventually reaches when the resistance of the medium prevents further acceleration. For example, a skydiver eventually stops accelerating and continues to fall at a steady speed.
In the case of the baseball, terminal velocity occurs when the drag force counterbalances the gravitational force, resulting in no net force and thus no acceleration.
  • At terminal velocity, \( F_d = F_g \)
  • Which implies: \( kv_t^2 = mg \)
  • This formula helps find the constant \( k \), using the known terminal speed \( v_t \).
So, when we talk about half terminal speed, this is the context we are looking at—a point where motion is slower, and forces are not yet balanced.
Newton's Second Law
Newton's Second Law is central to understanding the movement of any object. It tells us that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
  • Formula: Newton's Second Law can be expressed as: \( F = ma \)
  • This helps us calculate the acceleration, knowing the net force and mass.
In the baseball scenario, Newton's Second Law allows us to calculate the acceleration when the ball is moving up or down, considering the different directions and magnitudes of drag and gravitational forces.
By knowing the forces acting in each scenario, using this law helps determine how quickly the ball will slow down or speed up.

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Most popular questions from this chapter

A rock with mass \(m=3.00\) kg falls from rest in a viscous medium. The rock is acted on by a net constant downward force of 18.0 \(\mathrm{N}\) (a combination of gravity and the buoyant force exerted by the medium) and by a fluid resistance force \(f=k v\) where \(v\) is the speed in \(\mathrm{m} / \mathrm{s}\) and \(k=2.20 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}(\) see Section 5.3\() .\) (a) Find the initial acceleration \(a_{0 .}\) (b) Find the acceleration when the speed is 3.00 \(\mathrm{m} / \mathrm{s} .\) (c) Find the speed when the acceleration equals 0.1\(a_{0}\) (d) Find the terminal speed \(v_{\mathrm{t}}\) . (e) Find the coordinate, speed, and acceleration 2.00 s after the start of the motion. (f) Find the time required to reach a speed of 0.9\(v_{t} .\)

You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of 28.0 \(\mathrm{kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32,\) what magnitude of force must you apply?

A box with weight \(w\) is pulled at constant speed along a level floor by a force \(\vec{\boldsymbol{F}}\) that is at an angle \(\theta\) above the horizontal. The coefficient of kinetic friction between the floor and box is \(\mu_{\mathrm{k}}\) (a) In terms of \(\theta, \mu_{\mathrm{k}},\) and \(w\) calculate \(F .\) (b) For \(w=400 \mathrm{N}\) and \(\mu_{\mathrm{k}}=0.25,\) calculate \(F\) for \(\theta\) ranging from \(0^{\circ}\) to \(90^{\circ}\) in increments of \(10^{\circ} .\) Graph \(F\) versus \(\theta\) .(c) From the general expression in part (a), calculate the value of \(\theta\) for which the value of \(F\) , required to maintain constant speed, is a minimum. (Hint: At a point where a function is minimum, what are the first and second derivatives of the function? Here \(F\) is a function of \(\theta . )\) For the special case of \(w=400 \mathrm{N}\) and \(\mu_{\mathrm{k}}=0.25\) evaluate this optimal \(\theta\) and compare your result to the graph you constructed in part (b).

The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end (Fig. E5.46). Each arm supports a seat suspended from a cable 5.00 m long, the upper end of the cable being fastened to the arm at a point 3.00 \(\mathrm{m}\) from the central shaft. (a) Find the time of one revolution of the swing if the cable supporting a seat makes an angle of \(30.0^{\circ}\) with the vertical. (b) Does the angle depend on the weight of the passenger for a given rate of revolution?

A small car with mass 0.800 kg travels at constant speed on the inside of a track that is a vertical circle with radius 5.00 m (Fig. E5.42). If the normal force exerted by the track on the car when it is at the top of the track (point \(B\) is 6.00 \(\mathrm{N},\) what is the normal force on the car when it is at the bottom of the track (point \(A ) ?\)

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