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Chapter 7: Problem 2

How high will a \(1.85-\mathrm{kg}\) rock go if thrown straight up by someone who does \(80.0 \mathrm{~J}\) of work on it? Neglect air resistance.

Short Answer

Expert verified
The rock will rise to approximately 4.41 meters.

Step by step solution

01

Understanding Work and Gravitational Potential Energy

When work is done on an object, it is converted into kinetic energy which, in turn, gets converted into gravitational potential energy as the object rises. The rock's maximum height can be found using the principle of conservation of energy, where the work done on the rock becomes its gravitational potential energy at the maximum height.
02

Identify the Formula for Gravitational Potential Energy

The gravitational potential energy ( extit{PE}) at a height extit{h} is given by \[ PE = mgh, \]where \(m\) is the mass of the rock, \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)), and \(h\) is the height.
03

Set Work Done Equal to Potential Energy

Since all the work done on the rock is converted into potential energy at its maximum height, we set the work equal to the gravitational potential energy. Thus,\[80.0 \, \text{J} = 1.85 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot h.\]
04

Solve for Height

Rearrange the equation to solve for \(h\):\[ h = \frac{80.0}{1.85 \times 9.81}.\]Calculate the height:\[ h \approx \frac{80.0}{18.1485} \approx 4.41 \, \text{m}.\]The maximum height the rock will reach is approximately \(4.41 \, \text{m}\).

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is a way to understand the energy stored in an object due to its position in a gravitational field. Imagine holding a rock above the ground. It has potential energy because of its height. This type of energy depends on three key factors:
  • The object's mass (\(m\)) - Heavier objects store more potential energy.
  • The height (\(h\)) above the ground - The higher an object, the more potential energy it has.
  • The acceleration due to gravity (\(g\)) - Usually about \(9.81 \, \text{m/s}^2\) on Earth.
These elements combine in the formula: \(\text{PE} = mgh\). This means that when a rock is thrown upwards, it's converting its motion into potential energy, reaching a maximum just before it starts to fall back down.
Work-Energy Principle
The work-energy principle ties together the concept of work and energy. When someone works on an object, they transfer energy to it. This is crucial to understanding how things move. For example, when you throw a rock into the air, the work you do on the rock is stored as energy.
In mathematical terms, the work done (\(W\)) is equal to the change in energy:\[W = \Delta KE + \Delta PE.\] In this exercise, the work done, \(80.0 \, \text{J}\), is completely transformed into the rock's potential energy at its highest point. There, the kinetic energy (energy of motion) becomes zero because the rock momentarily stops before descending.
Kinetic Energy
Kinetic energy is all about motion. When the rock is first thrown, all the energy is kinetic because the rock is moving upwards. It can be calculated using the formula: \( KE = \frac{1}{2}mv^2 \)where \(m\) is the mass of the object and \(v\) is its velocity.
Initially, when you do work on the rock, this work is transferred into kinetic energy, propelling the rock upward. As the rock rises, its velocity decreases because gravitational force is opposing the motion. During this upward journey, kinetic energy is continually transformed into gravitational potential energy until the rock reaches its highest point. At that peak, all the initial kinetic energy is converted into gravitational potential energy, ready to turn back into kinetic energy as the rock begins to fall back down.

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Most popular questions from this chapter

(II) \(\mathrm { A } 46.0\) -kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225\(\mathrm { N }\) . For the first 11.0\(\mathrm { m }\) the floor is frictionless, and for the next 10.0\(\mathrm { m }\) the coefficient of friction is \(0.20 .\) What is the final speed of the crate after being pulled these 21.0\(\mathrm { m }\) ?

(1) How much work is required to stop an electron \(\left( m = 9.11 \times 10 ^ { - 31 } \mathrm { kg } \right)\) which is moving with a speed of \(1.40 \times 10 ^ { 6 } \mathrm { m } / \mathrm { s } ?\)

(III) An elevator cable breaks when a 925 -kg elevator is 22.5\(\mathrm { m }\) above the top of a huge spring \(( k =\) \(8.00 \times 10 ^ { 4 } \mathrm { N } / \mathrm { m } )\) at the bottom of the shaft. Calculate (a) the work done by gravity on the elevator before it hits the spring; \(( b )\) the speed of the elevator just betore striking the spring; \(( c )\) the amount the spring compresses (note that here work is done by both the spring and gravity.

A child is pulling a wagon down the sidewalk. For \(9.0 \mathrm{~m}\) the wagon stays on the sidewalk and the child pulls with a horizontal force of \(22 \mathrm{~N}\). Then one wheel of the wagon goes off on the grass so the child has to pull with a force of \(38 \mathrm{~N}\) at an angle of \(12^{\circ}\) to the side for the next \(5.0 \mathrm{~m}\). Finally the wagon gets back on the sidewalk so the child makes the rest of the trip, \(13.0 \mathrm{~m},\) with a force of \(22 \mathrm{~N}\). How much total work did the child do on the wagon?

What is the dot product of \(\overrightarrow{\mathbf{A}}=2.0 x^{2} \hat{\mathbf{i}}-4.0 x \hat{\mathbf{j}}+5.0 \hat{\mathbf{k}}\) and \(\overrightarrow{\mathbf{B}}=11.0 \hat{\mathbf{i}}+2.5 x \hat{\mathbf{j}}\) ?
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