/*! 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 93 A stone is thrown vertically upw... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 93

A stone is thrown vertically upward. On its way up it passes point \(A\) with speed \(v\), and point \(B, 3.00 \mathrm{~m}\) higher than \(A\), with speed \(\frac{1}{2} v .\) Calculate (a) the speed \(v\) and (b) the maximum height reached by the stone above point \(B\),

Short Answer

Expert verified
The speed \(v\) is approximately 8.85 m/s, and the maximum height above point B is 1 m.

Step by step solution

01

Understand the Energy Conservation Principle

The problem involves vertically thrown stones where kinetic energy and gravitational potential energy are exchanged during motion. For such problems, energy conservation can be a useful approach as the energy at point A and point B remains conserved, assuming no air resistance. Thus, the sum of kinetic energy and gravitational potential energy at any point during the stone's path is constant.
02

Express Kinetic and Potential Energies at Points A and B

At point A, the kinetic energy is \( \frac{1}{2} m v^2 \) and potential energy is \( mgh_A \). At point B, the kinetic energy is \( \frac{1}{2} m \left( \frac{1}{2}v \right)^2 \) and potential energy is \( mg(h_A + 3) \). These energies must be equal, thus: \[ \frac{1}{2} m v^2 + mgh_A = \frac{1}{2} m \left( \frac{1}{2}v \right)^2 + mg(h_A + 3) \]
03

Simplify and Solve the Energy Equation

Cancel the mass \(m\) from the equation as it is common in all terms: \[ \frac{1}{2} v^2 = \frac{1}{8}v^2 + g \times 3 \]Simplifying gives: \[ \frac{3}{8} v^2 = 3g \].Solving for \(v^2\) yields: \[ v^2 = 8g \].And solving for \(v\) gives: \[ v = \sqrt{8g} \].Substitute \(g = 9.8 \text{ m/s}^2\) to find \(v\).
04

Calculate the Speed \(v\)

Substituting \(g = 9.8 \text{ m/s}^2\), we have:\[ v = \sqrt{8 \times 9.8} \approx 8.85 \text{ m/s} \].
05

Calculate Maximum Height Above Point B

At the maximum height, the stone's speed is 0. Using energy conservation from point B to the highest point (point C), equate the kinetic energy at B to the potential energy difference between B and C:\[ \frac{1}{2} m \left( \frac{1}{2}v \right)^2 = mg(h_C - (h_A+3)) \]Solve for \(h_C - (h_A+3)\):\[ \frac{1}{8} v^2 = 9.8(h_C - (h_A+3)) \]Using \( v^2 = 8 \times 9.8 \), we have \( \frac{1}{8} \times 8 \times 9.8 \) simplifies to \( 9.8 \), then:\[ h_C - (h_A+3) = 1 \text{ m}\].Thus, the maximum height above B is 1 m.

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Key Concepts

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

Energy Conservation
Energy conservation is a powerful concept in physics that tells us energy cannot be created or destroyed. It can only change forms. In the case of the stone being thrown upwards, energy conservation simplifies the problem significantly. As it moves, the stone鈥檚 kinetic energy (energy of motion) transforms into gravitational potential energy (energy due to height).

At any point in its trajectory, the sum of kinetic energy (\(\frac{1}{2} mv^2\)) and potential energy (\(mgh\)) is constant, assuming no other external forces such as air resistance are involved. This consistent total energy allows us to predict the stone's behavior. For instance, at the highest point in its trajectory, all kinetic energy will have converted to potential energy, causing the stone to momentarily stop before falling back down. Understanding this principle can help solve various problems in physics relating to motion.
Kinetic Energy
Kinetic energy refers to the energy possessed by an object due to its motion. The formula to calculate kinetic energy is \(\frac{1}{2} mv^2\), where \(m\) is the mass and \(v\) is the velocity of the object.

In the stone problem, its kinetic energy determines how fast it is moving at any point in its path. For example, at point A, the stone has a certain kinetic energy due to its speed, \(v\). As the stone rises to point B, it slows down, reducing its kinetic energy to \(\frac{1}{2}v\). Because energy is being conserved, this loss in kinetic energy suggests a gain in potential energy.

This conversion continues until the stone reaches its highest point above B, where all its kinetic energy has been converted into potential energy, halting its upward motion until gravity forces it to come back down.
Potential Energy
Potential energy is the energy an object has due to its position or state. For instance, a stone held at a height has gravitational potential energy because of the Earth's gravity pulling on it. The formula is \(mgh\), where \(m\) is mass, \(g\) is the acceleration due to gravity, and \(h\) is the height above a reference point.

In this problem, as the stone rises from point A to point B, and further to its maximum height, its potential energy increases. The rise from A to B involves a height difference of 3 meters, leading to a notable increase in potential energy, which could be calculated with \(mg(h_A+3)\). When the stone reaches its peak above point B, the potential energy is at its maximum, and the kinetic energy is zero.

Understanding how potential energy varies with height helps in solving problems involving changes in vertical positions and the related energy transformations.

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Most popular questions from this chapter

A pilot flies horizontally at \(1300 \mathrm{~km} / \mathrm{h}\), at height \(h=35 \mathrm{~m}\) above initially level ground. However, at time \(t=0\), the pilot begins to fly over ground sloping upward at angle \(\theta=4.3^{\circ}\) (Fig. 2-38). If the pilot does not change the airplane's heading, at what time \(t\) does the plane strike the ground?

A parachutist bails out and freely falls \(50 \mathrm{~m}\). Then the parachute opens, and thereafter she decelerates at \(2.0 \mathrm{~m} / \mathrm{s}^{2}\). She reaches the ground with a speed of \(3.0 \mathrm{~m} / \mathrm{s}\). (a) How long is the parachutist in the air? (b) At what height does the fall begin?

A lead ball is dropped in a lake from a diving board \(5.20 \mathrm{~m}\) above the water. It hits the water with a certain velocity and then sinks to the bottom with this same constant velocity. It reaches the bottom \(4.80 \mathrm{~s}\) after it is dropped. (a) How deep is the lake? What are the (b) magnitude and (c) direction (up or down) of the average velocity of the ball for the entire fall? Suppose that all the water is drained from the lake. The ball is now thrown from the diving board so that it again reaches the bottom in \(4.80 \mathrm{~s}\). What are the (d) magnitude and (e) direction of the initial velocity of the ball?

A rock is dropped (from rest) from the top of a 60 -m-tall building. How far above the ground is the rock \(1.2 \mathrm{~s}\) before it reaches the ground?

In an arcade video game, a spot is programmed to move across the screen according to \(x=9.00 t-0.750 t^{3}\), where \(x\) is distance in centimeters measured from the left edge of the screen and \(t\) is time in seconds. When the spot reaches a screen edge, at either \(x=0\) or \(x=15,0 \mathrm{~cm}, t\) is reset to 0 and the spot starts moving again according to \(x(t) .\) (a) At what time after starting is the spot instantaneously at rest? (b) At what value of \(x\) does this occur? (c) What is the spot's acceleration (including sign) when this occurs? (d) Is it moving right or left just prior to coming to rest? (e) Just after? (f) At what time \(t>0\) does it first reach an edge of the screen?
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