/*! 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 61 A ball of mass \(m=0.75 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A ball of mass \(m=0.75 \mathrm{kg}\) is thrown straight upward with an initial speed of \(8.9 \mathrm{m} / \mathrm{s}\). (a) Plot the gravitational potential energy of the block from its launch height, \(y=0,\) to the height \(y=5.0 \mathrm{m} .\) Let \(U=0\) correspond to \(y=0\) (b) Determine the turning point (maximum height) of this mass.

Short Answer

Expert verified
(a) The plot shows potential energy increasing linearly with height. (b) The turning point is approximately 4.04 meters.

Step by step solution

01

Define the potential energy equation

Gravitational potential energy \( U \) as a function of height \( y \) is given by the equation:\[U = mgy\]where \( m = 0.75 \mathrm{kg} \) is the mass, \( g = 9.8 \mathrm{m/s^2} \) is the acceleration due to gravity, and \( y \) is the height.
02

Calculate potential energy at different heights

We will calculate the gravitational potential energy at various heights from \( y = 0 \) to \( y = 5.0 \mathrm{m} \). Substituting into the potential energy equation:- At \( y = 0 \): \( U = 0.75 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times 0 = 0 \, \mathrm{J} \)- At \( y = 1 \): \( U = 0.75 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times 1 = 7.35 \, \mathrm{J} \)- At \( y = 2 \): \( U = 0.75 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times 2 = 14.7 \, \mathrm{J} \)- At \( y = 3 \): \( U = 0.75 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times 3 = 22.05 \, \mathrm{J} \)- At \( y = 4 \): \( U = 0.75 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times 4 = 29.4 \, \mathrm{J} \)- At \( y = 5 \): \( U = 0.75 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} \times 5 = 36.75 \, \mathrm{J} \)
03

Understand the implication of U=0

Since we defined \( U = 0 \) at \( y = 0 \), this implies that the potential energy is 0 at the starting point. As the ball rises against gravity, its potential energy increases linearly with height.
04

Determine maximum height using energy conservation

Using the principle of conservation of energy, the initial kinetic energy is converted into gravitational potential energy at the maximum height. The equation is: \[\frac{1}{2}mv^2 = mgy_{max}\]Substitute \( m = 0.75 \, \mathrm{kg} \), \( g = 9.8 \, \mathrm{m/s^2} \), and \( v = 8.9 \, \mathrm{m/s} \):\[\frac{1}{2}(0.75)(8.9)^2 = (0.75)(9.8)y_{max}\] Solve for \( y_{max} \).
05

Solve for maximum height

Simplify the equation from the previous step:\[\frac{1}{2} \times 0.75 \times 79.21 = 7.35y_{max}\]\[29.705 = 7.35y_{max}\]\[y_{max} = \frac{29.705}{7.35} \approx 4.04 \, \mathrm{m}\]
06

Plot the potential energy against height

Using the calculated potential energies from Step 2, the plot should show potential energy on the vertical axis and height \( y \) on the horizontal axis. The plot should be a straight line starting at (0, 0) and going up to (5, 36.75) without any breaks.

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 91Ó°ÊÓ!

Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. When a ball is thrown straight upwards, it initially possesses kinetic energy due to its speed. The formula to calculate kinetic energy (KE) of an object is:\[KE = \frac{1}{2}mv^2\] where:
  • \(m\) is the mass of the object
  • \(v\) is the velocity of the object
For our specific example, the initial velocity is given as 8.9 m/s, and the mass is 0.75 kg. Plugging these values into the kinetic energy equation gives us the object's initial kinetic energy.When the ball is thrown upward, the kinetic energy begins to decrease as the object's height increases. This decrease is due to the conversion of kinetic energy into potential energy.In essence, the energy transformation is what enables the ball to reach its maximum height before it starts descending back toward the ground.
Energy Conservation
The law of energy conservation is a fundamental concept in physics. It states that energy can neither be created nor destroyed, only transformed from one form to another. In the context of our exercise, the total mechanical energy of the system—the sum of kinetic and potential energy—remains constant, provided no external forces like air resistance act on the ball.At the beginning of the motion, the ball has maximum kinetic energy and zero potential energy. As the ball rises, the kinetic energy converts into gravitational potential energy. The conversion continues until the ball reaches its maximum height, where its velocity becomes zero. At this point, all initial kinetic energy has been converted into gravitational potential energy.The mathematical expression for energy conservation is:\[\text{Initial Kinetic Energy} = \text{Potential Energy at maximum height}\]On solving for maximum height using the principle of conservation of energy, it is essential to ensure that all dimensions and units align correctly to get an accurate answer. The transition from kinetic to potential energy can be fully understood by following the energy transformation process throughout the projectile's journey.
Projectile Motion
Projectile motion occurs when an object is thrown into the air and moves under the influence of gravity alone, without any propulsion. In this exercise, throwing the ball upwards demonstrates simple one-dimensional projectile motion. In projectile motion, there are two key phases to consider:
  • Ascent: The ball goes up, decelerating because gravity opposes its motion upward.
  • Descent: After reaching the peak, the ball starts descending, reversing its motion direction and accelerating back down due to gravity.
Projectile motion is characterized by symmetric paths—the time it takes for the ball to reach the maximum height is the same as the time it takes to fall back down to the same level. It's essential to calculate the maximum height to determine how high the projectile goes before it begins its descent. This is where the concept of converting initial kinetic energy into potential energy comes into play. By carefully observing and plotting the projectile’s path, one can gain a deeper understanding of how forces like gravity influence an object in motion. The fundamental principles of projectile motion illustrate how different types of energy interplay, showcasing the beauty and predictability of physical laws.

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

A 1.40-kg block slides with a speed of 0.950 m/s on a frictionless horizontal surface until it encounters a spring with a force constant of \(734 \mathrm{N} / \mathrm{m}\). The block comes to rest after compressing the spring \(4.15 \mathrm{cm}\). Find the spring potential energy, \(U\), the kinetic energy of the block, \(K,\) and the total mechanical energy of the system, \(E\), for compressions of (a) \(0 \mathrm{cm}\) (b) \(1.00 \mathrm{cm}\) (c) \(2.00 \mathrm{cm}\) (d) \(3.00 \mathrm{cm},\) and (e) \(4.00 \mathrm{cm}\)

A pendulum bob with a mass of 0.13 kg is attached to a string with a length of \(0.95 \mathrm{m}\). We choose the potential energy to be zero when the string makes an angle of \(90^{\circ}\) with the vertical. (a) Find the potential energy of this system when the string makes an angle of \(45^{\circ}\) with the vertical. (b) Is the magnitude of the change in potential energy from an angle of \(90^{\circ}\) to \(45^{\circ}\) greater than, less than, or the same as the magnitude of the change from \(45^{\circ}\) to \(0^{\circ}\) ? Explain. (c) Calculate the potential energy of the system when the string is vertical.

Tension at the Bottom A ball of mass \(m\) is attached to a string of length \(L\) and released from rest at the point \(A\) in Figure \(8-29 .\) (a) Show that the tension in the string when the ball reaches point \(\mathrm{B}\) is \(3 \mathrm{mg}\), independent of the length \(l\). (b) Give a detailed physical explanation for the fact that the tension at point \(\mathrm{B}\) is independent of the length \(l.\)

Running Shoes The soles of a popular make of running shoe have a force constant of \(2.0 \times 10^{5} \mathrm{N} / \mathrm{m} .\) Treat the soles as ideal springs for the following questions. (a) If a \(62-\mathrm{kg}\) person stands in a pair of these shoes, with her weight distributed equally on both feet, how much does she compress the soles? (b) How much energy is stored in the soles of her shoes when she's standing?

A rock is thrown vertically upward from the top of a cliff that is \(32 \mathrm{m}\) high. When it hits the ground at the base of the cliff, the rock has a speed of \(29 \mathrm{m} / \mathrm{s}\). Assuming that air resistance can be ignored, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.

See all solutions

Recommended explanations on Physics Textbooks

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.