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What is the speed of a 0.15-kg baseball whose kinetic energy is 77 J?

Short Answer

Expert verified
The baseball's speed is approximately 32.03 m/s.

Step by step solution

01

Understand Kinetic Energy Formula

The kinetic energy (KE) of an object is calculated using the formula: \[ KE = \frac{1}{2}mv^2 \] where \( m \) is the mass of the object in kilograms, and \( v \) is its velocity in meters per second (m/s). We have \( KE = 77 \text{ J} \) and \( m = 0.15 \text{ kg} \). Our goal is to find the speed \( v \).
02

Rearrange the Formula

Rearrange the kinetic energy equation to solve for velocity \( v \). Multiply both sides of the equation by 2 to get rid of the fraction:\[ 2 \cdot KE = mv^2 \]Then divide both sides by \( m \) to isolate \( v^2 \): \[ v^2 = \frac{2 \cdot KE}{m} \]
03

Substitute the Known Values

Substitute the known values into the rearranged equation. \[ v^2 = \frac{2 \cdot 77}{0.15} \] First, calculate \( 2 \cdot 77 = 154 \). Then \[ v^2 = \frac{154}{0.15} \] \[ v^2 = 1026.67 \]
04

Solve for Velocity

Take the square root of both sides to solve for \( v \): \[ v = \sqrt{1026.67} \] Using a calculator, find that \[ v \approx 32.03 \text{ m/s} \].

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Key Concepts

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

Velocity
Velocity is an essential concept when discussing kinetic energy. It tells us the rate at which an object changes its position. In the context of physics, it's a vector, which means it has both magnitude and direction.
Unlike speed, which is scalar and only has magnitude, velocity helps us understand the direction as well. When solving problems involving kinetic energy, we're often looking for the velocity of an object, as is the case in our exercise.
To find velocity from kinetic energy, we rearrange the kinetic energy formula to isolate the velocity term:
  • First, multiply the kinetic energy by 2 to remove the fraction in the formula.
  • Then, divide this result by the object's mass.
  • Finally, take the square root of this division to solve for velocity.
This sequence leads us to understand how energy is distributed in the movement of an object. In our example, we calculated that the baseball is moving at approximately 32.03 m/s. This velocity indicates not only how fast the baseball is going, but it also plays a critical role in determining its kinetic energy.
Mass
Mass plays a crucial role in calculating kinetic energy. It's the measure of the amount of material in an object and is commonly measured in kilograms (kg).
In physics, mass is a fundamental property that influences how an object moves and reacts to forces. It's pivotal in kinetic energy equations, directly affecting the kinetic energy value. As the mass of an object increases, assuming constant velocity, the kinetic energy will proportionally increase too. This is because the formula for kinetic energy is directly dependent on mass.
In our exercise, we have a baseball with a mass of 0.15 kg. By incorporating mass into the kinetic energy formula and rearranging it to solve for velocity, we are reminded of the fundamental impact mass has on kinetic energy. The baseball's mass is relatively small, yet its kinetic energy is significantly sizeable, showing us that notwithstanding a modest mass, a high velocity contributes significantly to its energy status.
Energy Calculation
Energy calculation involves understanding the relationship between mass, velocity, and kinetic energy. It allows us to quantify the energy an object possesses due to its motion.
The formula for calculating kinetic energy is \( KE = \frac{1}{2}mv^2 \), where:
  • \( KE \) is kinetic energy in joules (J),
  • \( m \) is mass in kilograms (kg),
  • \( v \) is velocity in meters per second (m/s).
To perform an energy calculation, one may either use known values of mass and velocity to find kinetic energy or use kinetic energy and mass to solve for velocity.
In solving such problems, remembering to first isolate the variable of interest is key. For example, if velocity is unknown, rearrange the formula to solve for \( v \). This process is illustrated in the exercise where the baseball’s speed was calculated using given kinetic energy and mass values. Hence, energy calculation strategies are not only exclusive to theoretical exercises but also applicable to real-world scenarios, enhancing our understanding of motion dynamics.

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Most popular questions from this chapter

An object has a speed of \(3.5 \mathrm{~m} / \mathrm{s}\) and a kinetic energy of \(14 \mathrm{~J}\) at \(t=0\). At \(t=5.0 \mathrm{~s}\) the object has a speed of \(4.7 \mathrm{~m} / \mathrm{s}\). (a) What is the mass of the object? (b) What is the kinetic energy of the object at \(t=5.0 \mathrm{~s}\) ? (c) How much work was done on the object between \(t=0\) and \(t=5.0 \mathrm{~s}\) ?

A dog lifts a \(0.75-\mathrm{kg}\) bone straight up through a distance of \(0.11 \mathrm{~m}\). How much work was done by the dog?

How many joules of energy are in a kilowatt-hour?

Predict \(\&\) Explain You throw a ball upward and let it fall to the ground. Your friend drops an identical ball straight down to the ground from the same height. (a) Is the change in kinetic energy (from just after the ball is released until just before it hits the ground) of your ball greater than, less than, or equal to the change in kinetic energy of your friend's ball? (b) Choose the best explanation from among the following: C. The change in gravitational potential energy is the same for each ball, which means that the change in kinetic energy must also be the same. A. Your friend's ball converts all of its initial energy into kinetic energy. B. Your ball is in the air longer, which results in a greater change in kinetic energy.

Think \& Calculate A \(1100-\mathrm{kg}\) car is coasting on a horizontal road with a speed of \(19 \mathrm{~m} / \mathrm{s}\). After passing over an unpaved, sandy stretch \(32 \mathrm{~m}\) long, the car's speed has decreased to \(12 \mathrm{~m} / \mathrm{s}\). (a) Was the net work done on the car positive, negative, or zero? Explain. (b) Find the magnitude of the average net force on the car in the sandy section of the road.

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