/*! 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 24 A soccer ball with mass 0.420 \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 24

A soccer ball with mass 0.420 \(\mathrm{kg}\) is initially moving with speed 2.00 \(\mathrm{m} / \mathrm{s}\) . A soccer player kicks the ball, exerting a constant force of magnitude 40.0 \(\mathrm{N}\) in the same direction as the ball's motion. Over what distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 \(\mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
The distance is approximately 0.1365 meters.

Step by step solution

01

Identify Given Values

First, let's write down what is given in the problem. The mass of the soccer ball, \( m = 0.420 \, \mathrm{kg} \). The initial velocity of the ball, \( v_i = 2.00 \, \mathrm{m/s} \). The final velocity of the ball, \( v_f = 6.00 \, \mathrm{m/s} \). The force applied by the player, \( F = 40.0 \, \mathrm{N} \). We need to find the distance over which this force acts, \( d \).
02

Use Kinetic Energy Formula

The work done by the force changes the kinetic energy of the soccer ball. Use the kinetic energy formula. The change in kinetic energy, \( \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \). Substitute the given velocities and mass to find \( \Delta KE \).
03

Calculate Change in Kinetic Energy

Substitute the values into the kinetic energy formula: \( \Delta KE = \frac{1}{2} \cdot 0.420 \cdot (6.00)^2 - \frac{1}{2} \cdot 0.420 \cdot (2.00)^2 \). Calculate that \( \Delta KE = 6.30 - 0.84 = 5.46 \, \mathrm{J} \).
04

Use Work-Energy Principle

According to the work-energy principle, the work done on an object is equal to the change in its kinetic energy. The work done by the force is given by \( W = F \cdot d \). Therefore, \( F \cdot d = \Delta KE \).
05

Solve for Distance

Now, solve for \( d \) by rearranging the equation to \( d = \frac{\Delta KE}{F} \). Substitute in \( \Delta KE = 5.46 \, \mathrm{J} \) and \( F = 40.0 \, \mathrm{N} \). Thus, \( d = \frac{5.46}{40.0} \approx 0.1365 \, \mathrm{m} \).
06

Conclusion

The player's foot must be in contact with the ball over a distance of approximately 0.1365 meters to achieve the desired increase in speed.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 due to its motion. It depends on two main factors: the mass of the object and its velocity. The formula to calculate kinetic energy is given by \( KE = \frac{1}{2} m v^2 \), where \( m \) represents mass and \( v \) is velocity. This formula highlights that kinetic energy increases with both greater mass and higher velocity.
  • Mass: For a soccer ball with a mass of 0.420 kg, we can see that its mass is directly involved in calculating kinetic energy.
  • Velocity: The initial velocity of 2.00 m/s and the final velocity of 6.00 m/s play crucial roles in determining the change in kinetic energy.
Understanding that kinetic energy changes as velocity changes helps us solve problems involving movements, like in our soccer ball example.
Force
Force is any interaction that, when unopposed, will change the motion of an object. It can make an object move, change direction, or stop. In physics, force is measured in Newtons (N) and is often represented by the symbol \( F \).
  • In the soccer ball example, the player exerts a constant force of 40 N in the direction of the ball's motion.
  • This force is responsible for increasing the ball's velocity, which in turn affects its kinetic energy.
The work-energy principle states that the work (force applied over a distance) on an object results in a change in energy, specifically kinetic energy in this context. Understanding how to calculate force and its effects is essential in solving motion-related problems.
Mass
Mass is a measure of how much matter is in an object. It鈥檚 a fundamental property that does not change regardless of location or speed. In our example, the soccer ball has a mass of 0.420 kg, which is crucial for calculating kinetic energy and understanding the effect of force.
  • Mass directly affects motion; a more massive object requires more force to change its velocity than a lighter object.
  • In the kinetic energy formula, mass is directly proportional to the energy, meaning more mass results in more kinetic energy.
Knowing the mass of objects in motion helps us predict how they will behave when forces are applied. This is valuable for solving physics problems and understanding everyday phenomena like sports and vehicle movements.
Velocity
Velocity is a vector quantity that refers to the speed of an object in a particular direction. Unlike speed, which only measures how fast an object is moving, velocity tells us both how fast and in which direction.
  • The soccer ball in our problem starts with an initial velocity of 2.00 m/s and needs to reach a velocity of 6.00 m/s.
  • The change in velocity contributes to the change in kinetic energy, as seen in the equation \( \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) \).
Understanding velocity is crucial in determining how the motion of an object changes over time. When solving problems involving motion, accurately determining initial and final velocities helps us find how forces are applied and energy exchanged.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

CALC Varying Coefficient of Friction. A box is sliding with a speed of 4.50 \(\mathrm{m} / \mathrm{s}\) on a horizontal surface when, at point \(P\) it encounters a rough section. On the rough section, the coefficient of friction is not constant, but starts at 0.100 at \(P\) and increases linearly with distance past \(P\) , reaching a value of 0.600 at 12.5 \(\mathrm{m}\) past point \(P .\) (a) Use the work-energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid if the friction coefficient didn't increase but instead had the constant value of 0.100\(?\)

A 1.50 -kg book is sliding along a rough horizontal surface. At point \(A\) it is moving at \(3.21 \mathrm{m} / \mathrm{s},\) and at point \(B\) it has slowed to 1.25 \(\mathrm{m} / \mathrm{s}\) (a) How much work was done on the book between \(A\) and \(B ?\) (b) If \(-0.750 \mathrm{J}\) of work is done on the book from \(B\) to \(C\) , how fast is it moving at point \(C ?\) (c) How fast would it be moving at \(C\) if \(+0.750 \mathrm{J}\) of work were done on it from \(B\) to \(C\) ?

Six diesel units in series can provide 13.4 \(\mathrm{MW}\) of power to the lead car of a freight train. The diesel units have total mass \(1.10 \times 10^{6} \mathrm{kg}\) . The average car in the train has mass \(8.2 \times 10^{4} \mathrm{kg}\) and requires a horizontal pull of 2.8 \(\mathrm{kN}\) to move at a constant 27 \(\mathrm{m} / \mathrm{s}\) on level tracks. (a) How many cars can be in the train under these conditions? (b) This would leave no power for accelerating or climbing hills. Show that the extra force needed to accelerate the train is about the same for a \(0.10-\mathrm{m} / \mathrm{s}^{2}\) acceleration or a 1.0\(\%\) slope (slope angle \(\alpha=\arctan 0.010 )\) . (c) With the 1.0\(\%\) slope, show that an extra 2.9 \(\mathrm{MW}\) of power is needed to maintain the \(27-\mathrm{m} / \mathrm{s}\) speed of the diesel units. (d) With 2.9 \(\mathrm{MW}\) less power available, how many cars can the six diesel units pull up a 1.0\(\%\) slope at a constant 27 \(\mathrm{m} / \mathrm{s} ?\)

The spring of a spring gun has force constant \(k=400 \mathrm{N} / \mathrm{m}\) and negligible mass. The spring is compressed \(6.00 \mathrm{cm},\) and a ball with mass 0.0300 \(\mathrm{kg}\) is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 \(\mathrm{cm}\) long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal. (a) Calculate the speed with which the ball leaves the barrel if you can ignore friction. (b) Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 \(\mathrm{N}\) acts on the ball as it moves along the barrel. (c) For the situation in part (b), at what position along the barrel does the ball have the greatest speed, and what is that speed? (In this case, the maximum speed does not occur at the end of the barrel.)

A physics professor is pushed up a ramp inclined upward at \(30.0^{\circ}\) above the horizontal al as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and chair is 85.0 \(\mathrm{kg} .\) He is pushed 2.50 \(\mathrm{m}\) along the incline by a group of students who together exert a constant horizontal force of 600 \(\mathrm{N} .\) The professor's speed at the bottom of the ramp is 2.00 \(\mathrm{m} / \mathrm{s} .\) Use the work-energy theorem to find his speed at the top of the ramp.
See all solutions

Recommended explanations on Physics Textbooks

Quantum Physics

Read Explanation

Electromagnetism

Read Explanation

Translational Dynamics

Read Explanation

Fundamentals of Physics

Read Explanation

Oscillations

Read Explanation

Space Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.