/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 A baseball has a mass of 0.145 k... [FREE SOLUTION] | 91Ó°ÊÓ

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A baseball has a mass of 0.145 kg. (a) In batting practice a batter hits a ball that is sitting at rest on top of a post. The ball leaves the post with a horizontal speed of \(30.0 \mathrm{~m} / \mathrm{s}\). How much work did the force applied by the bat do on the ball? (b) During a game the same batter swings at a ball thrown by the pitcher and hits a line drive. Just before the ball is hit it is traveling at a speed of \(20.0 \mathrm{~m} / \mathrm{s},\) and just after it is hit it is traveling in the opposite direction at a speed of \(30.0 \mathrm{~m} / \mathrm{s}\). Whatis the total work done on the baseball by the force exerted by the bat? (c) How do the results of parts (a) and (b) compare? Explain.

Short Answer

Expert verified
The work done on the baseball by the bat in part (a) is 65.25 J, and in part (b) it is -87.75 J, indicating that the bat did more work in part (b) to reverse the direction of the ball.

Step by step solution

01

Calculate the work done on the ball in part (a)

In part (a), the ball was initially at rest, so the initial kinetic energy was 0, then it got a final velocity of 30.0 m/s. To calculate the work done we use \( W = \Delta KE = \frac{1}{2}m(v_{final}^{2} - v_{initial}^{2}) = \frac{1}{2}\times 0.145 kg \times (30.0 m/s)^2 = 65.25 J \)
02

Calculate the work done on the ball in part (b)

In part (b), the ball was initially moving with a speed of 20.0 m/s and final speed was -30.0 m/s (opposite direction means negative velocity). Again use the same formula: \( W = \Delta KE = \frac{1}{2}m(v_{final}^{2} - v_{initial}^{2}) = \frac{1}{2}\times 0.145 kg \times [(-30.0 m/s)^2 - (20.0 m/s)^2] = -87.75 J \) Note that the work done is negative indicating that the work was done against the direction of motion.
03

Compare the results of part (a) and (b)

In part (a) the work done on the baseball by the bat was positive, meaning the bat did work to increase the kinetic energy of the ball. While in part (b), the work done was negative as the bat did work to decrease the kinetic energy of the ball (or effectively reversed its direction). Hence the amount of work done by the bat was more in part (b) as compared to part (a).

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Key Concepts

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

Understanding Kinetic Energy
Kinetic energy is a key concept in physics, especially when dealing with moving objects like baseballs. It refers to the energy that an object possesses due to its motion. The faster something moves, the more kinetic energy it has. Kinetic energy can be calculated using the formula: \[KE = \frac{1}{2} mv^2\]Where:
  • \( KE \) stands for kinetic energy
  • \( m \) is the mass of the object, measured in kilograms
  • \( v \) is the velocity of the object, measured in meters per second
In the original exercise, the baseball's kinetic energy changes as the bat applies a force. Initially, in part (a), the ball has no kinetic energy because it is at rest. When hit, its kinetic energy rises as it speeds up. In part (b), even though the ball is already moving, it still gains more kinetic energy post-hit, despite a change in direction. This showcases how kinetic energy can be used to understand the energy involved in familiar situations like baseball games.
Physics Problem Solving Skills
Solving physics problems often involves using core concepts and formulas effectively. The work-energy theorem is pivotal for solving many mechanics problems. It states that the work done by a force on an object equals the change in kinetic energy of the object. This theorem gives us:\[W = \Delta KE = \frac{1}{2} m(v_{final}^{2} - v_{initial}^{2})\]To solve parts (a) and (b) of the exercise, this formula is key.
  • In part (a), since the ball starts at rest, the initial kinetic energy is zero.
  • For part (b), the negative sign in the velocity indicates the direction change, providing a complete picture of the energy shifts involved.
Physics problem-solving requires attention to units, arithmetic accuracy, and logical sequencing of steps. A clear understanding of underlying concepts like the work-energy theorem helps greatly. Identifying initial and final states and calculating changes in kinetic energy are common procedures in tackling mechanics problems in physics.
Applying Concepts in Mechanics
Mechanics is a branch of physics dealing with motion and forces. The interaction of forces and the resulting motion is at the heart of the mechanics study. In the given exercise, mechanics principles are applied to analyze the work done on a baseball. When a force acts on an object, changing its speed or direction, work is performed. This work can either increase or decrease the object's kinetic energy.
  • In part (a), the force of the bat adds kinetic energy, propelling the ball forward.
  • In part (b), the direction change results in a decrease in kinetic energy, illustrating the force applied in the opposite direction of the ball's initial motion.
By comparing these results, we gain insight into the effects of force and movement direction in sports scenarios. Understanding how forces interact in mechanics enriches the comprehension of everyday occurrences. Mechanics not only solves abstract problems but also explains the physics of real-world events, from sports to transportation.

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Most popular questions from this chapter

Using a cable with a tension of \(1350 \mathrm{~N}\), a tow truck pulls a car \(5.00 \mathrm{~km}\) along a horizontal roadway. (a) How much work does the cable do on the car if it pulls horizontally? If it pulls at \(35.0^{\circ}\) above the horizontal? (b) How much work does the cable do on the tow truck in both cases of part (a)? (c) How much work does gravity do on the car in part (a)?

A \(2.50 \mathrm{~kg}\) textbook is forced against one end of a horizontal spring of negligible mass that is fixed at the other end and has force constant \(250 \mathrm{~N} / \mathrm{m}\), compressing the spring a distance of 0.250 m. When released, the textbook slides on a horizontal tabletop with coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.30 .\) Use the work-energy theorem to find how far the textbook moves from its initial position before it comes to rest.

A luggage handler pulls a \(20.0 \mathrm{~kg}\) suitcase up a ramp inclined at \(32.0^{\circ}\) above the horizontal by a force \(\vec{F}\) of magnitude \(160 \mathrm{~N}\) that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is \(\mu_{\mathrm{k}}=0.300 .\) If the suitcase travels \(3.80 \mathrm{~m}\) along the ramp, calculate (a) the work done on the suitcase by \(\overrightarrow{\boldsymbol{F}} ;\) (b) the work done on the suitcase by the gravitational force; (c) the work done on the suitcase by the normal force; (d) the work done on the suitcase by the friction force; (e) the total work done on the suitcase. (f) If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled \(3.80 \mathrm{~m}\) along the ramp?

Meteor Crater. About 50,000 years ago, a meteor crashed into the earth near present-day Flagstaff, Arizona. Measurements from 2005 estimate that this meteor had a mass of about \(1.4 \times 10^{8} \mathrm{~kg}\) (around 150,000 tons) and hit the ground at a speed of \(12 \mathrm{~km} / \mathrm{s}\). (a) How much kinetic energy did this meteor deliver to the ground? (b) How does this energy compare to the energy released by a 1.0 megaton nuclear bomb? (A megaton bomb releases the same amount of energy as a million tons of TNT, and 1.0 ton of TNT releases \(4.184 \times 10^{9} \mathrm{~J}\) of energy.)

You and your bicycle have combined mass \(80.0 \mathrm{~kg}\). When you reach the base of a bridge, you are traveling along the road at \(5.00 \mathrm{~m} / \mathrm{s}\) (Fig. \(\mathrm{P} 6.74\) ). At the top of the bridge, you have climbed a vertical distance of \(5.20 \mathrm{~m}\) and slowed to \(1.50 \mathrm{~m} / \mathrm{s}\). Ignore work done by friction and any inefficiency in the bike or your legs. (a) What is the total work done on you and your bicycle when you go from the base to the top of the bridge? (b) How much work have you done with the force you apply to the pedals?

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