/*! 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 58 A pitcher throws a \(0.140-kg\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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A pitcher throws a \(0.140-kg\) baseball, and it approaches the bat at a speed of 40.0 \(m/s\) . The bat does \(W_{\mathrm{nc}}=70.0 J\) of work on the ball in hitting it. Ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is 25.0 \(m\) above the point of impact.

Short Answer

Expert verified
The speed of the ball is approximately 45.93 m/s.

Step by step solution

01

Identify Given Values

We know the mass of the baseball is \(m = 0.140\, \text{kg}\), the initial speed of the ball is \(v_i = 40.0\, \text{m/s}\), the work done on the ball by the bat is \(W_{\text{nc}} = 70.0\, \text{J}\), and the height \(h = 25.0\, \text{m}\) above the point of impact.
02

Apply the Work-Energy Principle

According to the work-energy principle, the work done on an object is equal to the change in its mechanical energy. This can be written as:\[ W_{\text{nc}} = \Delta KE + \Delta PE \]where \(\Delta KE\) is the change in kinetic energy and \(\Delta PE\) is the change in potential energy.
03

Calculate Change in Kinetic Energy

The change in kinetic energy \(\Delta KE\) can be expressed as:\[ \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) \]where \(v_f\) is the final velocity we want to find.
04

Calculate Change in Potential Energy

The potential energy changes as the ball moves upwards, and this can be calculated using:\[ \Delta PE = mgh \]where \(g = 9.8\, \text{m/s}^2\) is the acceleration due to gravity.
05

Substitute Known Values into the Equation

Substitute the known values into the work-energy equation:\[ 70 = \frac{1}{2} \cdot 0.140 \cdot (v_f^2 - 40^2) + 0.140 \cdot 9.8 \cdot 25 \]
06

Simplify the Equation

Calculate \(\Delta PE\):\[ \Delta PE = 0.140 \cdot 9.8 \cdot 25 = 34.3 \text{ J} \]Substitute this into the equation:\[ 70 = \frac{1}{2} \cdot 0.140 \cdot (v_f^2 - 1600) + 34.3 \]
07

Solve for Final Velocity

Rearrange and solve for \(v_f^2\):1. Subtract \(34.3\) from both sides: \[ 35.7 = \frac{1}{2} \cdot 0.140 \cdot (v_f^2 - 1600) \]2. Multiply both sides by \(2/0.140\): \[ 510 = v_f^2 - 1600 \]3. Add \(1600\) to both sides: \[ v_f^2 = 2110 \]4. Take the square root of both sides: \[ v_f = \sqrt{2110} \approx 45.93 \text{ m/s} \]
08

Conclusion: Final Answer

The speed of the ball after it leaves the bat and rises 25.0 m above the point of impact is approximately \(45.93 \text{ m/s}\).

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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 possessed by an object due to its motion. It is a fundamental concept in understanding how energy is transformed and conserved in physics. The kinetic energy (KE) of an object can be calculated using the formula:
  • \( KE = \frac{1}{2} m v^2 \)
Here, \(m\) is the mass of the object, and \(v\) is its velocity. In this exercise, the baseball has an initial speed, which means it has initial kinetic energy. When the bat hits the ball, it does work on it, changing its kinetic energy. Understanding this change is crucial for solving problems involving motion and forces. In our example, the kinetic energy must be recalculated after an additional 70 J of work by the bat, further changing the velocity of the ball as it moves upward.
Potential Energy
Potential energy refers to the energy stored in an object due to its position or state. For objects close to Earth, gravitational potential energy (PE) is the most common form. It is determined by the object's height above a reference point and can be calculated using the formula:
  • \( PE = mgh \)
Where \(m\) is the mass, \(g\) is the acceleration due to gravity (approximately \(9.8 \text{ m/s}^2\)), and \(h\) is the height above a reference point. In our baseball scenario, as the ball rises to 25 meters after being hit, its potential energy increases. This increase in potential energy comes at the expense of its kinetic energy, affecting the final velocity of the ball post-impact. Hence, understanding potential energy is key to solving such physics problems, where energy transformations between potential and kinetic forms are involved.
Mechanical Energy
Mechanical energy is the sum of kinetic energy and potential energy in a system. It represents the total energy available for doing work. The work-energy principle states that when work is done on an object, it affects its mechanical energy. The relationship is given by:
  • \( W_{\text{nc}} = \Delta KE + \Delta PE \)
Where \(W_{\text{nc}}\) is the work done by non-conservative forces, \(\Delta KE\) is the change in kinetic energy, and \(\Delta PE\) is the change in potential energy. Understanding mechanical energy allows one to see how energy shifts between its kinetic and potential forms. In this scenario, the baseball's mechanical energy changes as it moves upwards and the bat applies additional work. By calculating how the work done by the bat alters both kinetic and potential energies, we can find the final speed of the ball.
Physics Problem Solving
Physics problem-solving often involves breaking down complex scenarios into manageable parts by applying fundamental principles, such as the conservation of energy. Here's a basic approach to solving problems similar to the baseball exercise:
  • Identify the variables: Determine all the known quantities such as mass, initial velocity, height, and work done.
  • Apply relevant formulas: Use the work-energy principle, kinetic energy formula, and potential energy formula to set up equations.
  • Solve step by step: Substitute known values into the equations, simplify, and solve for the unknown.
  • Check your work: Verify the calculations by checking if they make physical sense.
By methodically following these steps, you can systematically approach and solve a wide range of physics problems. Understanding how energy is conserved and transformed in different scenarios helps in accurately finding solutions, as demonstrated in the bat and ball exercise.

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Most popular questions from this chapter

A 2.00-kg rock is released from rest at a height of 20.0 m. Ignore air resistance and determine the kinetic energy, gravitational potential energy, and total mechanical energy at each of the following heights: 20.0, 10.0, and 0 m.

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