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You throw a ball straight up into the air. It reaches a maximum height and returns to your hand. At what location(s) is the kinetic energy of the ball (a) a maximum and (b) a minimum? At what location(s) is the potential energy of the ball (c) a maximum and (d) a minimum?

Short Answer

Expert verified
(a) Kinetic energy is a maximum at the start and end of the throw, (b) minimum at maximum height. (c) Potential energy is maximum at maximum height, (d) minimum at the start and end of the throw.

Step by step solution

01

Understanding Maximum and Minimum Kinetic Energy

Kinetic Energy (KE) is maximum when the velocity of the ball is greatest. This occurs at the start of the throw (at the point of release) and just before it is caught (when it returns to your hand). The ball has its highest velocity at these two points due to the initial throw and as it returns to the same height, making KE maximum at both the starting and ending points of the throw.
02

Locating Minimum Kinetic Energy

Kinetic Energy (KE) is minimum when the velocity of the ball is zero. This occurs at the maximum height of the ball's trajectory because gravity has decelerated the ball's upward motion to zero velocity before it begins to descend. Thus, the KE is minimum at the maximum height.
03

Identifying Maximum Potential Energy

Potential Energy (PE) is related to the height of the ball above the ground. It reaches a maximum when the ball is at its highest point because this is when it has the greatest distance from the ground. Therefore, the maximum potential energy occurs at the maximum height.
04

Determining Minimum Potential Energy

Potential Energy (PE) is minimum when the ball's height above the ground is minimum. This is at the point of release and the point of return (when the ball is in your hand), as these points are at the same, lowest vertical level in this throw-return scenario.

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Key Concepts

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

Ball Trajectory
When you throw a ball straight up, its path or "trajectory" is initially an upward curve before reaching a peak and coming back down. This motion results in a parabolic arc. Understanding this trajectory is essential for solving many problems involving projectile motion and energy concepts.
  • Initially, the ball moves up against the force of gravity, which causes it to slow down over time.
  • At its peak or apex, the ball momentarily stops before changing direction to descend back down.
  • This path is mostly vertical in a simple throw but can have a longer horizontal path in more complex situations.
By analyzing the ball's trajectory, you can better understand where the kinetic and potential energies are maximized or minimized.
Velocity
Velocity refers to the speed of an object in a specific direction. For the thrown ball, this concept is crucial in understanding energy changes.
  • As the ball is thrown upwards, its velocity decreases due to the opposing force of gravity until it reaches zero at the maximum height.
  • Upon descent, velocity increases as gravity pulls the ball back towards the ground.
  • Velocity is highest when the ball is released and when it is caught, meaning these are the points of maximum kinetic energy.
Grasping the concept of velocity helps in predicting the behavior of the ball at various points in its path.
Maximum Height
The maximum height of the ball is a critical point in its trajectory. This is where several important changes happen:
  • The ball's upward velocity becomes zero as it stops rising and prepares to fall.
  • Kinetic energy here is at a minimum since the velocity is zero.
  • Conversely, potential energy is at its peak because the ball is farthest from the ground.
This highest point not only marks the turnaround in motion but also the pivotal exchange between potential and kinetic energies. Understanding maximum height is crucial in solving energy-related problems.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field, often associated with its height above the ground.
  • The higher the ball is, the greater its potential energy. This is why GPE is highest at the maximum height of the ball.
  • Conversely, GPE is least when the ball is at or near your hand, at the point of release and when it returns, due to its minimum height.
  • GPE contributes to the ball's total mechanical energy, balancing with kinetic energy during the ball's flight.
A solid understanding of gravitational potential energy enhances comprehension of how potential and kinetic energies interplay in a moving object.

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Most popular questions from this chapter

Analyze Engine 1 produces twice the power of engine 2 . Is it correct to conclude that engine 1 does twice as much work as engine 2? Explain.

A small motor runs a lift that raises a load of bricks weighing 836 N to a height of 10.7 m in 23.2 s. Assuming that the bricks are lifted with constant speed, what is the minimum power the motor must produce?

A small motor runs a lift that raises a load of bricks weighing \(836 \mathrm{~N}\) to a height of \(10.7 \mathrm{~m}\) in \(23.2 \mathrm{~s}\). Assuming that the bricks are lifted with constant speed, what is the minimum power the motor must produce?

Assess System 1 has a force of \(10 \mathrm{~N}\) and a speed of \(5 \mathrm{~m} / \mathrm{s}\). System 2 has a force of \(20 \mathrm{~N}\) and a speed of \(2 \mathrm{~m} / \mathrm{s}\). Which system has the greater power? Explain.

Human-Powered Flight Human-powered aircraft require a pilot to pedal, as on a bicycle, and to produce a sustained power output of about \(0.30 \mathrm{hp}(1 \mathrm{hp}=746 \mathrm{~W})\). The Gossamer Albatross flew across the English Channel on June 12,1979 , in \(2 \mathrm{~h} 49 \mathrm{~min}\). (a) How much energy did the pilot expend during the flight? (b) How many candy bars (280 Cal per bar) would the pilot have to consume to be "fueled up" for the flight? Note that a nutritional calorie (1 Cal) is equivalent to 1000 calories (1000 cal) as defined in physics. In addition, the conversion factor between calories and joules is as follows: \(1 \mathrm{Cal}=1000 \mathrm{cal}=1 \mathrm{kcal}=4186 \mathrm{~J} .\)

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