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Chapter 6: Problem 13

Animal Energy. BIO Adult cheetahs, the fastest of the great cats, have a mass of about 70 \(\mathrm{kg}\) and have been clocked running at up to 72 \(\mathrm{mph}(32 \mathrm{m} / \mathrm{s})\) . (a) How many joules of kinetic energy does such a swift cheetah have? (b) By what factor would its kinetic energy change if its speed were doubled?

Short Answer

Expert verified
(a) 35840 J; (b) The kinetic energy quadruples if speed is doubled.

Step by step solution

01

Write the formula for kinetic energy

The formula for kinetic energy \( KE \) of an object with mass \( m \) moving with velocity \( v \) is given by:\[KE = \frac{1}{2} m v^2\]This formula will be used to calculate the kinetic energy of the cheetah.
02

Calculate the kinetic energy of the cheetah

Given that the mass \( m \) of the cheetah is \( 70 \, \text{kg} \) and its speed \( v \) is \( 32 \, \text{m/s} \), we substitute these values into the kinetic energy formula:\[KE = \frac{1}{2} \times 70 \, \text{kg} \times (32 \, \text{m/s})^2\]Calculate the expression:\[KE = \frac{1}{2} \times 70 \times 1024 = 35 \times 1024 = 35840 \, \text{Joules}\]
03

Determine how kinetic energy changes with doubled speed

To find out how the kinetic energy changes if the speed is doubled, we start by letting the new speed be \( 2v \). Substitute this new velocity into the kinetic energy formula:\[KE_{new} = \frac{1}{2} m (2v)^2 = \frac{1}{2} m \times 4v^2\]This means the kinetic energy becomes:\[KE_{new} = 4 \times \left(\frac{1}{2} m v^2\right) = 4 \times KE\]Thus, the cheetah's kinetic energy is multiplied by 4 when its speed is doubled.

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Key Concepts

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

Physics Problems
Solving physics problems involves understanding the principles and laws that govern the physical world. In the case of kinetic energy, it鈥檚 important to start with identifying the given data points, such as the mass and velocity of an object, and to clearly understand what the problem asks you to find.
  • First, read the problem carefully to extract all the necessary details. Here, the mass of the cheetah is 70 kg and its speed is 32 m/s.
  • Then, determine what you need to calculate鈥攊n this instance, the kinetic energy at this speed and how it changes with speed variations.
Having a systematic approach to physics problems can make the process more manageable and less daunting. By breaking down a problem into smaller, sequential steps, you can tackle even complex questions with ease.
Energy Calculation
Energy calculations, particularly for kinetic energy, involve applying the known formula to the problem at hand. Let's delve into calculating the kinetic energy of the cheetah. The formula for kinetic energy is expressed as:\[ KE = \frac{1}{2} m v^2 \]
  • Insert the provided values: the mass \( m = 70 \, \text{kg} \) and the velocity \( v = 32 \, \text{m/s} \).
  • Perform the calculation as follows: \[ KE = \frac{1}{2} \times 70 \, \text{kg} \times (32 \, \text{m/s})^2 \]
  • First calculate the square of the velocity: \((32)^2 = 1024\).
  • Then multiply by the mass and the constant: \(35 \times 1024 = 35840 \, \text{Joules} \).
This step confirms the amount of kinetic energy the cheetah possesses at a speed of 32 m/s. Properly understanding these calculations strengthens your grasp on how kinetic energy varies with mass and velocity.
Formula Application
Applying formulas correctly is crucial in physics when exploring relationships such as between velocity and kinetic energy. In the exercise, you explored how changing the cheetah's speed influences its kinetic energy.
  • When speed is doubled, assume the new speed is \(2v\). Substitute into the formula: \[ KE_{new} = \frac{1}{2} m (2v)^2 = \frac{1}{2} m \times 4v^2 \]
  • This transformation shows that the kinetic energy is multiplied by a factor of four.
  • Using the original kinetic energy value, you can further demonstrate this increase: \[ KE_{new} = 4 \times KE \]
Understanding the relationship between speed and kinetic energy through formulas deepens your comprehension of how physical changes impact energy, preparing you for more intricate physics challenges.

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Most popular questions from this chapter

A Near-Earth Asteroid. On April \(13,2029\) (Friday the 13th!), the asteroid 99942 Apophis will pass within \(18,600\) mi of the earth- about \(\frac{1}{13}\) the distance to the moon! It has a density of \(2600 \mathrm{kg} / \mathrm{m}^{3},\) can be modeled as a sphere 320 \(\mathrm{m}\) in diameter, and will be traveling at 12.6 \(\mathrm{km} / \mathrm{s}\) . (a) If, due to a small disturbance in its orbit, the asteroid were to hit the earth, how much kinetic energy would it deliver? (b) The largest nuclear bomb ever tested by the United States was the "Castle/Bravo" bomb, having a yield of 15 megatons of TNT. (A megaton of TNT releases \(4.184 \times 10^{15}\) J of energy.) How many Castle/Bravo bombs would be equivalent to the energy of Apophis?

How many joules of energy does a \(100-\) watt light bulb use per hour? How fast would a 70 -kg person have to run to have that amount of kinetic energy?

Pushing a Cat. Your cat "Ms." (mass 7.00 kg) is trying to make it to the top of a frictionless ramp 2.00 \(\mathrm{m}\) long and inclined upward at \(30.0^{\circ}\) above the horizontal. Since the poor cat can't get any traction on the ramp, you push her up the entire length of the ramp by exerting a constant \(100-\mathrm{N}\) force parallel to the ramp. If Ms. takes a running start so that she is moving at 2.40 \(\mathrm{m} / \mathrm{s}\) at the bottom of the ramp, what is her speed when she reaches the top of the incline? Use the work-energy theorem.

BIO 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 \(\mathrm{W}\) W). Which choice would burn up more energy, and how much energy (in joules) would it burn? Why is it that the more intense exercise actually burns up less energy than the less intense exercise?

CALC The gravitational pull of the earth on an object is inversely proportional to the square of the distance of the object from the center of the earth. At the earth's surface this force is equal to the object's normal weight \(m g,\) where \(g=9.8 \mathrm{m} / \mathrm{s}^{2},\) and at large distances, the force is zero. If a \(20,000-\mathrm{kg}\) asteroid falls to earth from a very great distance away, what will be its minimum speed as it strikes the earth's surface, and how much kinetic energy will it impart to our planet? You can ignore the effects of the earth's atmosphere.
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