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Chapter 8: Problem 56

(II) A meteorite has a speed of \(90.0 \mathrm{~m} / \mathrm{s}\) when \(850 \mathrm{~km}\) above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in \(3.25 \mathrm{~m}\). (a) What is its speed just before striking the sand? (b) How much work does the sand do to stop the meteorite (mass \(=575 \mathrm{~kg}\) )? (c) What is the average force exerted by the sand on the meteorite? (d) How much thermal energy is produced?

Short Answer

Expert verified
Approximate speed before impact: 4091 m/s. Work done by sand: -4,790,878,878.75 J. Average force by sand: -1474168796.38 N. Thermal energy: 4,790,878,878.75 J.

Step by step solution

01

Determine the Final Speed Before Impact

The initial speed of the meteorite when it's 850 km above Earth is 90.0 m/s. We use the formula for gravitational potential energy to find the speed just before striking the sand. The conversion from height to velocity can be calculated using energy conservation. The change in potential energy is converted to kinetic energy:\[ v^2 = u^2 + 2gs \]where \( u = 90 \text{ m/s} \), \( s = 850,000 \text{ m} \), \( g = 9.81 \text{ m/s}^2 \).Now substitute the values:\[ v^2 = 90^2 + 2(9.81)(850,000) \]Calculate \( v \):\[ v = \sqrt{90^2 + 2 \times 9.81 \times 850,000} \approx 4091 \text{ m/s} \].Thus, the speed just before striking the sand is approximately 4091 m/s.
02

Calculate the Work Done by the Sand

The work done by the sand to stop the meteorite can be found using the work-energy principle. The work done is equal to the change in kinetic energy of the meteorite.\[ \text{Work} = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \]where \( m = 575 \text{ kg} \), \( v = 0 \text{ m/s} \), \( u = 4091 \text{ m/s} \).Substitute the known values:\[ \text{Work} = \frac{1}{2} \times 575 \times 4091^2 - 0 \]Calculate the work done:\[ \text{Work} = 0 - \frac{1}{2} \times 575 \times 4091^2 = -4790878878.75 \text{ J} \].The sand does \(-4,790,878,878.75 \text{ J}\) of work to stop the meteorite.
03

Determine the Average Force Exerted by the Sand

The average force exerted by the sand can be calculated using the work done and the distance over which the force acts.\[ \text{Force} = \frac{\text{Work}}{\text{distance}} \]where the distance is \(3.25 \text{ m} \) and the work is \(-4790878878.75 \text{ J}\).\[ \text{Force} = \frac{-4790878878.75}{3.25} \]Calculate the force:\[ \text{Force} \approx -1474168796.38 \text{ N} \].The average force is approximately \(-1474168796.38 \text{ N}\) (the negative sign indicates direction).
04

Calculate the Thermal Energy Produced

The thermal energy produced is equal to the work done by the sand as all the kinetic energy is converted into thermal energy.The thermal energy produced is \( 4,790,878,878.75 \text{ J} \), which is the magnitude of the work done to bring the meteorite to rest.

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Key Concepts

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

Kinetic Energy
Understanding kinetic energy is crucial to solving problems related to moving objects. Kinetic energy is the energy an object possesses due to its motion. It is given by the formula \[ KE = \frac{1}{2}mv^2 \] where \( m \) is the mass of the object, and \( v \) is the velocity.
  • The greater the speed, the higher the kinetic energy.
  • Kinetic energy depends on the square of the velocity, which means doubling the speed quadruples the kinetic energy.
In the exercise, the meteorite's kinetic energy changes twice. First, as it accelerates towards Earth due to gravity. Then, when the kinetic energy is converted to other forms of energy as the meteorite is stopped by the sand.
Recognizing changes in kinetic energy helps in understanding how energies transform as the meteorite travels and eventually comes to rest.
Potential Energy
Potential energy is the stored energy based on position, height, or configuration. For objects near the Earth鈥檚 surface, gravitational potential energy can be calculated using:\[ PE = mgh \]where \( m \) is mass, \( g \) is gravity (\( 9.81 \, \text{m/s}^2 \)), and \( h \) is height above the ground.
  • Potential energy is higher when the object is higher above the ground.
  • Potential energy converts to kinetic energy as the object falls.
In the exercise, the meteorite has significant potential energy at 850 km above the Earth, which decreases as it falls, increasing its kinetic energy instead.
This conversion plays a critical role in determining the speed just before impact.
Average Force
The average force relates to how energy and movement are controlled over a particular distance or time. It is calculated using the formula:\[ F_{\text{avg}} = \frac{\text{Work done}}{\text{Distance}} \]
  • The force is the measure of how much the sand slowed the meteorite down over 3.25 meters.
  • A greater force implies more work was done over a shorter distance.
In this context, the negative sign of force suggests that the direction of force exerted by the sand is opposite to the meteorite鈥檚 motion.
Understanding average force provides insight into how quickly and efficiently a force can bring an object to rest. Force is essential in understanding impacts and collisions.
Thermal Energy
Thermal energy arises from the conversion of kinetic energy when objects and surfaces interact by contact. In the exercise, this is the energy generated as the meteorite comes to halt due to the sand's resistance.
  • All kinetic energy of the meteorite converts to thermal energy since it's brought to a stop.
  • Thermal energy is seen in the form of heat which may increase the temperature of both the meteorite and sand slightly.
This conversion is complete because the work done by the sand, through stopping the meteor, completely dissipates the initial kinetic energy into heat. It provides insight into energy transformation during interactions between surfaces.

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Most popular questions from this chapter

(I) Jane, looking for Tarzan, is running at top speed \((5.0 \mathrm{~m} / \mathrm{s})\) and grabs a vine hanging vertically from a tall tree in the jungle. How high can she swing upward? Does the length of the vine affect your answer?

(III) The potential energy of the two atoms in a diatomic (two-atom) molecule can be written $$ U(r)=-\frac{a}{r^{6}}+\frac{b}{r^{12}} $$ where \(r\) is the distance between the two atoms and \(a\) and \(b\) are positive constants. (a) At what values of \(r\) is \(U(r)\) a minimum? A maximum? (b) At what values of \(r\) is \(U(r)=0 ?\) (c) Plot \(U(r)\) as a function of \(r\) from \(r=0\) to \(r\) at a value large enough for all the features in \((a)\) and \((b)\) to show. ( \(d\) ) Describe the motion of one atom with respect to the second atom when \(E<0,\) and when \(E>0 .(e)\) Let \(F\) be the force one atom exerts on the other. For what values of \(r\) is \(F>0, F<0, F=0 ?\) (f) Determine \(F\) as a function of \(r\).

A particle moves where its potential energy is given by \(U(r)=U_{0}\left[\left(2 / r^{2}\right)-(1 / r)\right] .(a)\) Plot \(U(r)\) versus \(r .\) Where does the curve cross the \(U(r)=0\) axis? At what value of \(r\) does the minimum value of \(U(r)\) occur? (b) Suppose that the particle has an energy of \(E=-0.050 U_{0} .\) Sketch in the approximate turning points of the motion of the particle on your diagram. What is the maximum kinetic energy of the particle, and for what value of \(r\) does this occur?

(II) A particle is constrained to move in one dimension along the \(x\) axis and is acted upon by a force given by \(\overrightarrow{\mathbf{F}}(x)=-\frac{k}{x^{3}} \hat{\mathbf{i}}\) where \(k\) is a constant with units appropriate to the SI system. Find the potential energy function \(U(x)\), if \(U\) is arbitrarily defined to be zero at \(x=2.0 \mathrm{~m},\) so that \(U(2.0 \mathrm{~m})=0 .\)

(II) What should be the spring constant \(k\) of a spring designed to bring a \(1200-\mathrm{kg}\) car to rest from a speed of \(95 \mathrm{~km} / \mathrm{h}\) so that the occupants undergo a maximum acceleration of \(5.0 g ?\)
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