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Estimate the average force that a major-league pitcher exerts on the baseball when he throws a fastball. Express your answer in pounds. In your solution, list the quantities for which you estimate values and any assumptions you make. Do a web search to help determine the values you use in making your estimates.

Short Answer

Expert verified
The estimated average force, in pounds, that a major-league pitcher exerts on a baseball when throwing a fastball is approximately 3.64 pounds.

Step by step solution

01

Identify and Collect Variables

From a web search, it is known that a standard baseball has a mass of about 0.145 kg. For acceleration, knowing that many fastballs travel at about 100 miles per hour and that it takes roughly 0.4 seconds to reach home plate (18.4 meters away), use the formula \( a = \Delta v / t \) for constant acceleration to estimate the pitch's acceleration.
02

Convert Velocity

Before calculating the acceleration, first convert the velocity from miles per hour to meters per second, as we must use SI units for calculation. The conversion could be done as follows: \( 1 \text{ mile} = 1609.34 \text{ meters} \) and \( 1 \text{ hour} = 3600 \text{ seconds} \), so 100 miles per hour will be equal to \( 100 * 1609.34 / 3600 = 44.7 \text{ m/s} \).
03

Estimate Acceleration and Force

Now estimate the average acceleration knowing initial velocity (\( v_i = 0 \text{ m/s} \) as the ball starts from rest), final velocity (\( v_f = 44.7 \text{ m/s} \)) and time of 0.4 seconds. The acceleration can be calculated as \( a = \Delta v / t = v_f - v_i / t = 44.7 / 0.4 = 111.75 \text{ m/s}^2 \). Subsequently, the force exerted by the pitcher can be found using \( F = m * a = 0.145 * 111.75 = 16.2 \text{ Newtons} \).
04

Convert Force to Pounds

Finally, convert the force exerted by the pitcher from Newtons to pounds. The conversion could be done as follows: \( 1 \text{ Newton} = 0.2248 \text{ pounds} \), therefore, the force exerted by the pitcher will be \( 16.2 * 0.2248 = 3.64 \text{ pounds} \).

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Key Concepts

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

Acceleration
Acceleration is the rate at which an object's velocity changes over time. It's a vector quantity, meaning it has both magnitude and direction. When a baseball is thrown, its speed increases from zero to a certain value in a very short period of time, which is where acceleration comes into play.
To calculate acceleration (\( a \)), you need two primary pieces of information: the change in velocity (\( \Delta v \)) and the time (\( t \)) it takes for that change to occur. The formula used is: \[ a = \frac{\Delta v}{t} \]This formula shows that acceleration is directly related to the change in velocity and inversely related to the time period over which the change occurs.
For example, if a fastball goes from rest (0 \( \text{m/s} \)) to 44.7 \( \text{m/s} \) in 0.4 seconds, the acceleration would be 111.75 \( \text{m/s}^2 \). This significant acceleration is one reason why pitcher's arms are subject to high stress during the action of throwing.
Velocity Conversion
Before you can calculate acceleration in consistent units, it's essential to convert the velocity into an appropriate unit system. In scientific calculations, the International System of Units (SI) is often used, so converting speeds to meters per second (\( \text{m/s} \)) is a standard practice.
For example, if you have a velocity in miles per hour (\( \text{mph} \)), like 100 mph, you need to convert this into \( \text{m/s} \). This process involves two key conversions:
  • 1 mile equals 1609.34 meters.
  • 1 hour equals 3600 seconds.
Using these conversions, 100 mph converts to 44.7 \( \text{m/s} \).
This conversion is crucial as it ensures that all physical quantities, such as distance, time, and speed, are compatible within the same unit system, making calculations consistent and accurate.
Newton's Laws of Motion
Newton's Laws of Motion are fundamental principles that describe the behavior of objects in motion and are essential for understanding force calculation. The second law is particularly relevant when dealing with force and acceleration.
Newton’s Second Law states that the force (\( F \)) acting on an object is equal to the mass (\( m \)) of the object multiplied by its acceleration (\( a \)). Mathematically, this is expressed as: \[ F = m \cdot a \]This means that the force exerted on an object is directly proportional to the mass of the object and the acceleration it experiences.
In the context of a pitcher throwing a baseball, once you've calculated the baseball’s acceleration and know its mass, you can determine the average force the pitcher applies by using this law. This understanding not only helps in calculating the force accurately but also offers insight into how changes in mass or acceleration affect the force required.
Unit Conversion
Unit conversion is a critical step in physics calculations, including force calculation. It's the process of converting one unit of measure into another, facilitating consistent computations.
Converting units allows you to present your results in the most practical or requested format, which is particularly important in regions using different measurement systems.
Take, for instance, the conversion from Newtons to pounds when calculating force. This is necessary when the answer needs to be expressed in a form more familiar to the audience, or as requested.
  • 1 Newton equals 0.2248 pounds.
Therefore, if a calculation results in a force of 16.2 Newtons, it would convert to approximately 3.64 pounds.
By mastering unit conversions, you ensure that your results are not only mathematically correct but also tailored to the context required, making your findings more accessible to broader audiences.

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Most popular questions from this chapter

A box rests on a frozen pond, which serves as a frictionless horizontal surface. If a fisherman applies a horizontal force with magnitude \(48.0 \mathrm{~N}\) to the box and produces an acceleration of magnitude \(2.20 \mathrm{~m} / \mathrm{s}^{2}\), what is the mass of the box?

Starting at time \(t=0\), net force \(F_{1}\) is applied to an object that is initially at rest. (a) If the force remains constant with magnitude \(F_{1}\) while the object moves a distance \(d\), the final speed of the object is \(v_{1} .\) What is the final speed \(v_{2}\) (in terms of \(v_{1}\) ) if the net force is \(F_{2}=2 F_{1}\) and the object moves the same distance \(d\) while the force is being applied? (b) If the force \(F_{1}\) remains constant while it is applied for a time \(T,\) the final speed of the object is \(v_{1} .\) What is the final speed \(v_{2}\) (in terms of \(v_{1}\) ) if the applied force is \(F_{2}=2 F_{1}\) and is constant while it is applied for the same time \(T ?\) In a later chapter we'll call force times distance work and force times time impulse and associate work and impulse with the change in speed.)

The upward normal force exerted by the floor is \(620 \mathrm{~N}\) on an elevator passenger who weighs \(650 \mathrm{~N}\). What are the reaction forces to these two forces? Is the passenger accelerating? If so, what are the magnitude and direction of the acceleration?

An astronaut's pack weighs \(17.5 \mathrm{~N}\) when she is on the earth but only \(3.24 \mathrm{~N}\) when she is at the surface of a moon. (a) What is the acceleration due to gravity on this moon? (b) What is the mass of the pack on this moon?

You have just landed on Planet X. You release a \(100 \mathrm{~g}\) ball from rest from a height of \(10.0 \mathrm{~m}\) and measure that it takes \(3.40 \mathrm{~s}\) to reach the ground. Ignore any force on the ball from the atmosphere of the planet. How much does the \(100 \mathrm{~g}\) ball weigh on the surface of Planet X?

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