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Chapter 7: Problem 12

\(\cdot\) It takes 4.186 \(\mathrm{J}\) of energy to raise the temperature of 1.0 \(\mathrm{g}\) of water by \(1.0^{\circ} \mathrm{C}\) (a) How fast would a 2.0 \(\mathrm{g}\) cricket have to jump to have that much kinetic energy? (b) How fast would a 4.0 \(\mathrm{g}\) cricket have to jump to have the same amount of kinetic energy?

Short Answer

Expert verified
The 2.0 g cricket must jump at 64.7 m/s, while the 4.0 g cricket must jump at 45.7 m/s.

Step by step solution

01

Understanding Kinetic Energy

Kinetic energy (KE) is given by the formula: \( KE = \frac{1}{2}mv^2 \). Here, \( m \) is the mass of the object (in kg), and \( v \) is the velocity (in m/s). We need to find the velocity for which this energy equals 4.186 J.
02

Converting Mass

Firstly, convert the mass of the crickets from grams to kilograms since the standard unit for mass in physics is kilograms. For the 2.0 g cricket: \[ m_1 = 2.0 \text{ g} \times \frac{1}{1000} \text{ kg/g} = 0.002 \text{ kg} \]For the 4.0 g cricket:\[ m_2 = 4.0 \text{ g} \times \frac{1}{1000} \text{ kg/g} = 0.004 \text{ kg} \]
03

Solving for Velocity of 2.0g Cricket

Use the kinetic energy formula to solve for the velocity of the 2.0 g cricket:\[ 4.186 = \frac{1}{2} \cdot 0.002 \cdot v^2 \]First, isolate \( v^2 \):\[ v^2 = \frac{4.186}{0.001} \]\[ v^2 = 4186 \]Thus, the velocity \( v \) is:\[ v = \sqrt{4186} \approx 64.7 \text{ m/s} \]
04

Solving for Velocity of 4.0g Cricket

Use the kinetic energy formula to solve for the velocity of the 4.0 g cricket:\[ 4.186 = \frac{1}{2} \cdot 0.004 \cdot v^2 \]First, isolate \( v^2 \):\[ v^2 = \frac{4.186}{0.002} \]\[ v^2 = 2093 \]Thus, the velocity \( v \) is:\[ v = \sqrt{2093} \approx 45.7 \text{ m/s} \]

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Key Concepts

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

Physics Problems
Physics problems can be challenging, but they're an exciting way to apply mathematical formulas to real-world situations. When solving these problems, it's crucial to begin by understanding the variables and what the problem is asking you to find. In this example, we needed to calculate the velocity of crickets so that their kinetic energy would match a specific amount.
  • Identify the problem: Find out what you need to calculate and the given data.
  • Use the right formula: Here, we used the formula for kinetic energy, \( KE = \frac{1}{2}mv^2 \) where \( m \) is mass and \( v \) is velocity.
  • Unit conversion: Always convert your units into the standard form, such as kilograms for mass.
Key to solving physics problems is understanding these steps and applying them consistently. Make sure to double-check your calculations, especially when dealing with different units.
Energy Conversion
Energy conversion is the process of changing energy from one form to another. In our exercise, the focus was on converting kinetic energy to a numerical value using the kinetic energy formula. Kinetic energy is the energy an object possesses because of its motion. Understanding energy conversion can help in recognizing how various energy forms relate to each other.
Kinetic energy is dependent on both the mass and velocity of an object:
  • Mass (\( m \)): The amount of matter in the object, affecting how much kinetic energy the object can have.
  • Velocity (\( v \)): The speed at which the object is moving, playing a pivotal role since kinetic energy increases with the square of velocity \( (v^2) \).
The formula involves multiplying the mass and the square of the velocity, then halving it. This action shows how energy is converted from its potential form (the cricket poised to jump) to kinetic form (the cricket in flight). Understanding this principle is vital for many physics applications, from observing crickets to understanding massive celestial bodies.
Velocity Calculation
Calculating velocity requires manipulating the kinetic energy formula to solve for \( v \). Let's walk through how to determine the velocity required for our cricket example:
Firstly, convert mass to kilograms to fit the formula's requirements. For instance, convert 2.0 g to 0.002 kg.
Next, substitute the known values into the kinetic energy formula \( KE = \frac{1}{2}mv^2 \). With \( KE \) as 4.186 J and \( m \) as 0.002 kg:
  • Rearrange to solve for \( v^2 \): \[ v^2 = \frac{2 \times KE}{m} \]
  • Input known values: \[ v^2 = \frac{2 \times 4.186}{0.002} = 4186 \]
  • Calculate \( v \) by finding the square root of the result: \[ v = \sqrt{4186} \approx 64.7 \text{ m/s} \]
This calculation allows us to find the cricket's velocity needed for a specific amount of energy. Using mathematical manipulation, we effectively turned energy values into velocity, showcasing how physics allows for the conversion of energy into real-world speed.

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) All birds, independent of their size, must maintain a power output of \(10-25\) watts per kilogram of body mass in order to fly by flapping their wings. (a) The Andean giant hummingbird (Patagona gigas) has mass 70 \(\mathrm{g}\) and flaps its wings 10 times per second while hovering. Estimate the amount of work done by such a hummingbird in each wingbeat. (b) \(\mathrm{A} 70\) -kg athlete can maintain a power output of 1.4 \(\mathrm{kW}\) for no more than a few seconds; the steady power output of a typical athlete is only 500 \(\mathrm{W}\) or so. Is it possible for a human-powered aircraft to fly for extended periods by flapping its wings? Explain.

\(\bullet\) Rescue. Your friend (mass 65.0 \(\mathrm{kg}\) ) is standing on the ice in the middle of a frozen pond. There is very little friction between her feet and the ice, so she is unable to walk. Fortu- nately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for 3.00 s and accelerate your friend from rest to a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) while you remain at rest. What is the average power supplied by the force you applied?

\(\bullet\) \(\bullet\) Pebbles of weight \(w\) are launched from the edge of a verti- cal cliff of height \(h\) at speed \(v_{0} .\) How fast (in terms of the quan- tities just given) will these pebbles be moving when they reach the ground if they are launched (a) straight up, (b) straight down, (c) horizontally away from the cliff, and (d) at an angle \(\theta\) above the horizontal? (e) How would the answers to the pre- vious parts change if the pebbles weighed twice as much?

\(\bullet\) \(\bullet\) \(\bullet\) Automobile air-bag safety. An automobile air bag cush- ions the force on the driver in a head-on collision by absorbing her energy before she hits the steering wheel. Such a bag can be modeled as an elastic force, similar to that produced by a spring. (a) Use energy conser- vation to show that the effec- tive force constant \(k\) of the air bag is \(k=m v_{0}^{2} / x_{\text { max }}^{2},\) where \(m\) is the mass of the driver, \(v_{0}\) is the speed of the car (and driver) at the instant of the accident, and \(x_{\max }\) is the maxi- mum distance the bag gets compressed, which, in a severe accident, would be the dis- tance from the driver's body to the steering wheel. (b) Show that the maximum force the air bag would exert on the driver is \(F_{\max }=m v_{0}^{2} / x_{\max }\) (c) Now let's put in some realistic numbers. Experimental tests have shown that injury occurs when a force density greater than \(5.0 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}\) acts on human tissue. (The total force is this force density times the area over which it acts. As the accom- panying figure shows, the force of the air bag acts mostly on the upper front half of the driver's body, over an area of about 2500 \(\mathrm{cm}^{2} .\) (Check your own body to see if this is reasonable.) Use this value to calculate the total force on the driver's body at the threshold of injury. (d) Use your results to calculate the effective force constant \(k\) of the air bag and the maximum speed for which the bag will prevent injury to a 65 kg driver if she is 30 \(\mathrm{cm}\) from the steering wheel at the instant of impact. Express the speed in \(\mathrm{m} / \mathrm{s}\) and mph. (e) How could you design a safer air bag for higher speed collisions? What things could you alter to do this? Would it be safe to make a stiffer air bag by inflating it more? Explain your reasoning.

\(\bullet\) \(\bullet\) Energy requirements of the body. A 70 \(\mathrm{kg}\) human uses energy at the rate of \(80 \mathrm{J} / \mathrm{s},\) on average, for just resting and sleeping. When the person is engaged in more strenuous activities, the rate can be much higher. (a) If the individual did nothing but rest, how many food calories per day would she or he have to eat to make up for those used up? (b) In what forms is energy used when a person is resting or sleep- ing? In other words, what happens to those 80 \(\mathrm{J} / \mathrm{s} ?\) Hint: What kinds of energy, mechanical and otherwise, do our body components have?) (c) If an average person rested and did other low-level activity for 16 hours (which consumes 80 \(\mathrm{J} / \mathrm{s} )\) and did light activity on the job for 8 hours (which consumes \(200 \mathrm{J} / \mathrm{s} ),\) how many calories would she or he have to con- sume per day to make up for the energy used up?
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