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Chapter 7: Problem 17

\(\cdot\) A 0.145 kg baseball leaves a pitcher's hand at a speed of 32.0 \(\mathrm{m} / \mathrm{s} .\) If air drag is negligible, how much work has the pitcher done on the ball by throwing it?

Short Answer

Expert verified
The pitcher did 74.24 Joules of work on the baseball.

Step by step solution

01

Understand the Problem

We need to find the work done on the baseball by the pitcher as it leaves their hand at a given speed, 32.0 m/s. We are told that the mass of the baseball is 0.145 kg. We assume that all the work is used to increase the kinetic energy of the baseball, as air drag is negligible.
02

Identify the Relevant Formula

The work done by the pitcher is equal to the change in kinetic energy of the baseball. The formula for kinetic energy (KE) is given by:\[KE = \frac{1}{2}mv^2\]where \(m\) is the mass of the baseball and \(v\) is its velocity. Since the initial velocity \(v_0\) is 0 (assuming the baseball was at rest relative to the pitcher before being thrown), we can focus solely on its final kinetic energy when it leaves the pitcher's hand.
03

Calculate the Kinetic Energy

Substitute the given values into the kinetic energy formula:\[KE = \frac{1}{2} \times 0.145 \, \text{kg} \times (32.0 \, \text{m/s})^2\]This simplifies to:\[KE = 0.0725 \, \text{kg} \times 1024 \, \text{m}^2/\text{s}^2\]
04

Compute the Work Done

Calculate the kinetic energy using the numbers obtained in the previous step:\[KE = 0.0725 \, \text{kg} \times 1024 \, \text{m}^2/\text{s}^2 = 74.24 \, \text{Joules}\]Thus, the work done by the pitcher is 74.24 J, as it corresponds to the increase in kinetic energy of the baseball.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses because of its motion. It's a fundamental concept in physics that helps explain how energy is transferred and transformed.
Imagine you're moving. Everything you do with motion, from walking to swinging a bat, involves kinetic energy.
For the baseball example, when the pitcher throws the ball, the ball gradually picks up speed and thus kinetic energy.
  • The formula for kinetic energy of an object is given by: \[ KE = \frac{1}{2}mv^2 \] where \( m \) is mass and \( v \) is the velocity of the object.
  • In our example, the baseball's mass is 0.145 kg, and it's thrown at 32.0 m/s, yielding a certain amount of kinetic energy when it leaves the pitcher's hand.
  • This kinetic energy is what's acquired from the work done by the pitcher, leading us into the next key concept!
Work Done
Work done in physics is closely related to energy transfer. When a force causes an object to move, work is done. The amount of this work is determined by the force applied and distance covered by that object when that force is acting.
That sounds a bit technical, doesn't it? Let's break it down! Suppose you have a cart. When you push it, you're applying a force, and when it moves, there's work done on it.
In terms of equations, work done (W) can be represented as:
  • \[ W = F \cdot d \cdot \cos(\theta) \] where \( F \) is the force, \( d \) is the distance the object is moved, and \( \theta \) is the angle between the force and motion direction.
  • For our exercise, we are looking at work as the change in kinetic energy, which simplifies our task, as the formula for calculating work becomes \( W = \Delta KE \).
  • The work done by the pitcher equaled 74.24 Joules. This resulted solely from throwing the ball to achieve that specific speed of 32.0 m/s.
Physics Calculations
Physics calculations might seem complex initially, but they boil down to using the right formulas and correctly interpreting the given data.
Understanding physical principles, such as how energy changes and work operates, is crucial for calculating outcomes like those in the baseball exercise.
  • First, ensure all variables are well-defined. For instance, knowing the baseball's mass and velocity aids in calculating kinetic energy precisely.
  • Carefully substitute these values into the known physics equations step by step. Here, we had the formula for kinetic energy and used it to find the work done.
  • It's always beneficial to double-check units and calculations, ensuring consistency throughout the process. For this problem, ensuring joules as the final unit indicates it relates to energy.
  • Practicing these calculations for various scenarios helps concretize your understanding of physics principles.
Remember, approach each problem methodically, breaking it down into smaller, manageable parts, and physics will seem like much less of a complex puzzle!

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Most popular questions from this chapter

\(\bullet\) How high can we jump? The maximum height a typical human can jump from a crouched start is about 60 \(\mathrm{cm} .\) By how much does the gravitational potential energy increase for a 72 \(\mathrm{kg}\) person in such a jump? Where does this energy come from?

A tennis player hits a 58.0 g tennis ball so that it goes straight up and reaches a maximum height of 6.17 m. How much work does gravity do on the ball on the way up? On the way down?

\(\bullet\) \(\bullet\) At the site of a wind farm in North Dakota, the average wind speed is \(9.3 \mathrm{m} / \mathrm{s},\) and the average density of air is 1.2 \(\mathrm{kg} / \mathrm{m}^{3} .\) (a) Calculate how much kinetic energy the wind contains, per cubic meter, at this location. (b) No wind turbine can capture all of the energy contained in the wind, the main reason being that capturing all the energy would require stop- ping the wind completely, meaning that air would stop flowing through the turbine. Suppose a particular turbine has blades with a radius of 41 \(\mathrm{m}\) and is able to capture 35\(\%\) of the avail- able wind energy. What would be the power output of this tur- bine, under average wind conditions?

\(\bullet\) A tandem (two-person) bicycle team must overcome a force of 165 \(\mathrm{N}\) to maintain a speed of 9.00 \(\mathrm{m} / \mathrm{s}\) . Find the power required per rider, assuming that each contributes equally.

\(\bullet\) \(\bullet\) A typical flying insect applies an average force equal to twice its weight during each downward stroke while hovering. Take the mass of the insect to be \(10 \mathrm{g},\) and assume the wings move an average downward distance of 1.0 \(\mathrm{cm}\) during each stroke. Assuming 100 downward strokes per second, estimate the average power output of the insect.
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