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A baseball of mass 0.25 kg is hit at home plate with a speed of \(40 \mathrm{m} / \mathrm{s}\). When it lands in a seat in the left-field bleachers a horizontal distance \(120 \mathrm{m}\) from home plate, it is moving at \(30 \mathrm{m} / \mathrm{s}\). If the ball lands \(20 \mathrm{m}\) above the spot where it was hit, how much work is done on it by air resistance?

Short Answer

Expert verified
The work done by air resistance on the baseball is -37.5 J, indicating that air resistance acts against the motion of the baseball, causing it to lose energy.

Step by step solution

01

Break the initial velocity into components

Given the initial speed of 40 m/s, we need to find the horizontal and vertical components of the velocity. Let \(v_{0x}\) be the horizontal component and \(v_{0y}\) be the vertical component of the initial velocity. Since the ball has to travel a horizontal distance of 120 m and land 20 m above the starting point, we can use the formulas for projectile motion: \(D_x = v_{0x}t \) \(H = v_{0y}t - \frac{1}{2}gt^2\) Where \(D_x = 120\,\text{m}\): horizontal distance, \(H = 20\,\text{m}\): vertical distance, \(g= 9.8\,\text{m}/\text{s}^2\): acceleration due to gravity and, \(t\): time of travel. Note that we cannot determine the exact values of \(v_{0x}\) and \(v_{0y}\) yet since we have two variables and only one equation. We will continue to the next step.
02

Calculate initial kinetic energy

Calculate the initial kinetic energy of the baseball, using the mass and given speed: \(KE_i = \frac{1}{2}mv^2\) Where \(m=0.25\,\text{kg}\): mass of the baseball and \(v=40\,\text{m}/\text{s}\): initial speed of the baseball. \(KE_i = \frac{1}{2}(0.25\,\text{kg})(40\,\text{m}/\text{s})^2\) \(KE_i = 200 \,\text{J}\)
03

Calculate change in potential energy

Calculate the change in potential energy of the baseball when it reaches the final height: \(\Delta PE = mgh\) Where \(h =20 \,\text{m}\) is the height from the ground. \(\Delta PE = (0.25\,\text{kg})(9.8\,\text{m}/\text{s}^2)(20\,\text{m})\) \(\Delta PE = 49 \,\text{J}\)
04

Calculate final kinetic energy

Calculate the final kinetic energy of the baseball when it is moving at 30 m/s: \(KE_f = \frac{1}{2}mv_f^2\) Where \(v_f=30\,\text{m}/\text{s}\): final speed of the baseball. \(KE_f = \frac{1}{2}(0.25\,\text{kg})(30\,\text{m}/\text{s})^2\) \(KE_f = 112.5 \,\text{J}\)
05

Apply work-energy principle

Apply the work-energy principle, and find the work done by air resistance: \(W = \Delta KE + \Delta PE\) Where \(W\): work done by air resistance. Substitute the values of \(\Delta KE\) and \(\Delta PE\) from previous steps: \(W = (112.5 - 200) + 49\) \(W = -37.5 \,\text{J}\) The work done by air resistance on the baseball is -37.5 J. Since the work done is negative, it indicates that air resistance acts against the motion of the baseball, causing it to lose energy.

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Key Concepts

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

Projectile Motion
Projectile motion refers to the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. In the exercise provided, a baseball is hit and travels through the air, showcasing characteristic projectile motion.

The motion can be analyzed in two dimensions: horizontal and vertical. The horizontal motion occurs at a constant velocity, due to the absence of horizontal forces (ignoring air resistance for a moment). On the other hand, the vertical motion is influenced by gravity, causing a parabolic trajectory. The formulas mentioned in the problem use the initial speed broken down into its horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) components to determine how far (\(D_x\)) and how high (\(H\)) the ball will travel. These equations reflect the uniform motion horizontally and the uniformly accelerated motion vertically, which are key features of projectile motion.
Kinetic Energy Calculation
Kinetic energy (\(KE\)) represents the energy of motion. Any object that is moving has kinetic energy, which is calculated by the formula: \[ KE = \frac{1}{2}mv^2 \] where \(m\) is the mass of the moving object and \(v\) is its velocity.

To calculate the kinetic energy in the provided example, the mass of the baseball and its velocity are used to find its initial kinetic energy (\(KE_i\)) when hit and its final kinetic energy (\(KE_f\)) when caught. This concept is crucial because it helps us understand that as the baseball moves and slows down due to air resistance, its kinetic energy changes, which can be quantified through these calculations.
Potential Energy Change
Potential energy, in this context, specifically refers to gravitational potential energy (\(PE\)), which is energy stored by objects due to their position relative to Earth. The potential energy of an object increases with its height. The change in potential energy (\(\Delta PE\)) when an object is raised to a certain height is given by: \[ \Delta PE = mgh \] where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity, and \(h\) is the height above the starting point.

In the exercise, when the ball lands 20 meters above the point from which it was hit, this potential energy change is a gain, which must be factored in alongside kinetic energy changes to ascertain the total work done on the ball through its flight.
Work-Energy Principle
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. The concept is fundamental in physics and can be expressed as: \[ W = \Delta KE + \Delta PE \] where \(W\) is the work done on the object, \(\Delta KE\) is the change in kinetic energy, and \(\Delta PE\) is the change in potential energy. Work can be positive (adding energy to the object) or negative (removing energy from the object).

In this particular problem, the calculation shows that air resistance does negative work on the baseball, meaning it takes energy away from the ball, which is evident in the reduction of its speed from 40 m/s to 30 m/s. The negative sign in the work done by air resistance indicates that it acts opposite to the direction of the ball's motion.

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Most popular questions from this chapter

Consider a block of mass 0.200 kg attached to a spring of spring constant \(100 \mathrm{N} / \mathrm{m}\). The block is placed on a frictionless table, and the other end of the spring is attached to the wall so that the spring is level with the table. The block is then pushed in so that the spring is compressed by \(10.0 \mathrm{cm} .\) Find the speed of the block as it crosses (a) the point when the spring is not stretched, (b) \(5.00 \mathrm{cm}\) to the left of point in (a), and (c) \(5.00 \mathrm{cm}\) to the right of point in (a).

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A camera weighing \(10 \mathrm{N}\) falls from a small drone hovering \(20 \mathrm{m}\) overhead and enters free fall. What is the gravitational potential energy change of the camera from the drone to the ground if you take a reference point of (a) the ground being zero gravitational potential energy? (b) The drone being zero gravitational potential energy? What is the gravitational potential energy of the camera (c) before it falls from the drone and (d) after the camera lands on the ground if the reference point of zero gravitational potential energy is taken to be a second person looking out of a building \(30 \mathrm{m}\) from the ground?

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