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Chapter 6: Problem 24

You throw a \(3.00 \mathrm{~N}\) rock vertically into the air from ground level. You observe that when it is \(15.0 \mathrm{~m}\) above the ground, it is traveling at \(25.0 \mathrm{~m} / \mathrm{s}\) upward. Use the work-energy theorem to find (a) the rock's speed just as it left the ground and (b) its maximum height.

Short Answer

Expert verified
The rock's speed just as it left the ground is approximately 36.754m/s. The maximum height that the rock reaches is approximately 32.01m.

Step by step solution

01

Understanding Information and Applying the Work-Energy Principle for Initial Speed

Given that when the rock is 15.0m above the ground, it is moving upward with a velocity of 25.0m/s. The work done by gravity, \(W_{g}\), when the rock moves this distance can be calculated by \(W_{g}=mgh\), where \(m\) is mass, \(g\) is the acceleration due to gravity, and \(h\) is the height. The mass can be found by using the force \(F=m*g\). According to the work-energy theorem, this work is equal to the change in kinetic energy, i.e., \(W_{g}=\Delta KE = KE_{f} - KE_{i}\). The kinetic energy can be calculated by \(KE=\frac{1}{2}mv^2\), which, for the final (when the rock is 15m high) and initial (just released from the ground) states, can be denoted as \(KE_{f}\) and \(KE_{i}\), respectively.
02

Calculating the Initial Speed

Firstly calculate the mass from \(F=mg\), get \(m=F/g = 3N/9.8m/s^2 = 0.306kg\). Then substitute \(m\), \(g\), and \(h\) into the equation for work done by gravity \(W_g=mgh\) to find \(W_g=0.306kg*9.8m/s^2*15m=44.97J\). Then calculate the final kinetic energy \(KE_f=\frac{1}{2}mv^2=\frac{1}{2}*0.306kg*(25m/s)^2=96.125J\). Substituting \(W_{g}\) and \(KE_{f}\) into the work-energy theorem \(W_{g}= \Delta KE = KE_{f} - KE_{i}\), the initial kinetic energy \(KE_{i}=KE_{f}-W_g=96.125J-44.97J=51.155J\). Finally, substitute \(KE_i\) into the kinetic energy formula to calculate the speed \(KE_i=\frac{1}{2}mv_0^2\), the initial speed \(v_0=\sqrt{\frac{2KE_i}{m}}=\sqrt{\frac{2*51.155J}{0.306kg}}\approx36.754m/s\).
03

Applying the Work-Energy Principle for Maximum Height

When the rock reaches its highest point, the kinetic energy is zero (the speed is zero). However, the gravitational potential energy is at a maximum due to the height. Since no non-conservative forces are performing work (like friction or thrust), mechanical energy is conserved.
04

Calculating the Maximum Height

The initial gravitational potential energy is 0 m. The initial mechanical energy \(E_{i} = KE_{i} + PE_{i}\) will be equal to final mechanical energy \(E_{f} = KE_{f} + PE_{f}\). Here, \(KE_{f}=0\), \(PE_{i}=0\), and \(PE_{f}=mgh_{max}\), where \(h_{max}\) is the height at the top. Also, \(KE_{i}\) calculated previously is 51.155J. Thus, the equation becomes \(E_{i} = E_{f} \Rightarrow KE_{i} + PE_{i} = KE_{f} + PE_{f}\), or \(51.155J=0+mgh_{max}\). Solving for the height, get \(h_{max}= \frac{E_{i}}{mg}= \frac{51.155J}{0.306kg*9.8m/s^2}\approx17.01m\).
05

Adding the Initial Height to find the Total Height

The total maximum height is the sum of the initial 15m height and the just calculated height of 17.01m, yielding a total maximum height of approximately 32.01m.

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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. When you throw a rock in the air, like in the exercise, it gains kinetic energy. The formula to calculate kinetic energy is: \[ KE = \frac{1}{2}mv^2 \] where \( m \) is the mass of the object and \( v \) is its velocity. This equation shows us that kinetic energy is proportional to the mass of the object and the square of its velocity. This means that even a small increase in speed results in a relatively large increase in kinetic energy.
  • Example: As the rock travels upwards at 15 meters, it moves at 25 m/s. We calculated, using the formula, that its kinetic energy is 96.125 Joules.
  • Understanding kinetic energy is crucial in determining how fast the rock was when it left the ground (initial speed) since the energy transforms as the rock moves.
Adjusting the velocity helps us calculate the rock鈥檚 speed when it was initially thrown upwards, providing insights into the conservation and transformation of energy.
Potential Energy
Potential energy is the stored energy of an object due to its position relative to other objects. For objects in a gravitational field, like Earth, this energy depends on the object's height and mass. The gravitational potential energy (GPE) formula is: \[ PE = mgh \] where \( m \) is mass, \( g \) is acceleration due to gravity (\(9.8 \ m/s^2\)), and \( h \) is height.
  • In our rock example, potential energy is highest when it reaches its peak height because it's furthest from the Earth.
  • When the rock is at 15 meters, we calculate the work done against gravity as potential energy. This helps find the total mechanical energy available.
Gravitational potential energy grows with increased height, thereby playing a key role in finding how high the rock will eventually ascend.
Conservation of Mechanical Energy
The principle of conservation of mechanical energy is core to understanding energy transformations. It states that if no non-conservative forces (like friction) are doing work, the total mechanical energy in a system remains constant. Mechanical energy is the sum of kinetic and potential energy: \[ E_{total} = KE + PE \] This principle helps us solve for the maximum height of the rock in our exercise. As the rock moves upwards:
  • Its kinetic energy decreases while potential energy increases, but the sum of both remains constant.
  • At the highest point, the rock's speed is zero, so kinetic energy is zero, implying all mechanical energy is potential energy.
By applying this principle, we determine the maximum height by equating the initial kinetic energy to the maximum potential energy, thus finding the rock's total energy conversion from kinetic motion to stored gravitational potential energy. This illustrates how energy transitions between forms while the total remains constant, ensuring no loss or gain.

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Most popular questions from this chapter

Should You Walk or Run? It is \(5.0 \mathrm{~km}\) from your home to the physics lab. As part of your physical fitness program, you could run that distance at \(10 \mathrm{~km} / \mathrm{h}\) (which uses up energy at the rate of \(700 \mathrm{~W}\) ), or you could walk it leisurely at \(3.0 \mathrm{~km} / \mathrm{h}\) (which uses energy at \(290 \mathrm{~W}\) ). Which choice would burn up more energy, and how much energy (in joules) would it burn? Why does the more intense exercise burn up less energy than the less intense exercise?

A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from \(x=0\) to \(x=6.9 \mathrm{~m}\) as you apply a force with \(x\) -component \(F_{x}=-[20.0 \mathrm{~N}+(3.0 \mathrm{~N} / \mathrm{m}) x] .\) How much work does the force you apply do on the cow during this displacement?

Meteor Crater. About 50,000 years ago, a meteor crashed into the earth near present-day Flagstaff, Arizona. Measurements from 2005 estimate that this meteor had a mass of about \(1.4 \times 10^{8} \mathrm{~kg}\) (around 150,000 tons) and hit the ground at a speed of \(12 \mathrm{~km} / \mathrm{s}\). (a) How much kinetic energy did this meteor deliver to the ground? (b) How does this energy compare to the energy released by a 1.0 megaton nuclear bomb? (A megaton bomb releases the same amount of energy as a million tons of TNT, and 1.0 ton of TNT releases \(4.184 \times 10^{9} \mathrm{~J}\) of energy.)

A 12.0 kg package in a mail-sorting room slides \(2.00 \mathrm{~m}\) down a chute that is inclined at \(53.0^{\circ}\) below the horizontal. The coefficient of kinetic friction between the package and the chute's surface is 0.40 . Calculate the work done on the package by (a) friction, (b) gravity, and (c) the normal force. (d) What is the net work done on the package?

When a car is hit from behind, its passengers undergo sudden forward acceleration, which can cause a severe neck injury known as whiplash. During normal acceleration, the neck muscles play a large role in accelerating the head so that the bones are not injured. But during a very sudden acceleration, the muscles do not react immediately because they are flexible; most of the accelerating force is provided by the neck bones. Experiments have shown that these bones will fracture if they absorb more than \(8.0 \mathrm{~J}\) of energy. (a) If a car waiting at a stoplight is rear-ended in a collision that lasts for \(10.0 \mathrm{~ms}\) what is the greatest speed this car and its driver can reach without breaking neck bones if the driver's head has a mass of \(5.0 \mathrm{~kg}\) (which is about right for a \(70 \mathrm{~kg}\) person)? Express your answer in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{mi} / \mathrm{h}\). (b) What is the acceleration of the passengers during the collision in part (a), and how large a force is acting to accelerate their heads? Express the acceleration in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(\mathrm{g}\) 's.
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