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Use this method to find how the speed with which animals of similar shape can run up a hill depends on the size of the animal. Let \(L\) represent some characteristic length, such as the height or diameter of the animal. Assume that the maximum rate at which the animal can do work is proportional to the animal's surface area: \(P_{\max } \propto L^{2} .\) Set the maximum power output equal to the rate of increase of gravitational potential energy and determine how the speed \(v\) depends on \(L\).

Short Answer

Expert verified
The speed \(v\) depends inversely on the animal's size \(L\): \(v \propto L^{-1}\).

Step by step solution

01

Understanding Power and Work Energy

The problem states that the animal's maximum power output is proportional to its surface area: \(P_{\max} \propto L^2\). Power, being the rate of doing work, can be written as \(P = \frac{dW}{dt}\) where \(dW\) is the work done in time \(dt\). We know that gravitational potential energy (GPE) for an animal running uphill is given by \(mgh\), where \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is height. To find \(v\), we equate this to \(P_{\max}\).
02

Relating Mass to Volume and Length

Assume the animal has density \(\rho\), then its mass \(m\) can be expressed in terms of volume \(V\), assuming the animal is roughly spherical or cuboidal: \(m = \rho L^3\) since \(V \propto L^3\).
03

Expressing Power in Terms of Height and Speed

We rewrite the relationship between power and speed. Knowing that GPE change per unit time is \(mgh\), for an incline, this can be reformed to \(mgv\): \(P_{\max} = mgv\). Substitute \(m = \rho L^3\) to express power relative to length and speed: \(P_{\max} = \rho g v L^3 \).
04

Combining Proportional Relationships

Using the proportional relationship \(P_{\max} \propto L^2\), we substitute it into \(P_{\max} = \rho g v L^3\) to obtain \(v \propto \frac{L^2}{L^3} = L^{-1}\). This means speed \(v\) is inversely proportional to length \(L\).
05

Conclusion about Speed's Dependence on Length

Combining all the derived relationships, it becomes clear that the running speed of the animal decreases as the size (length \(L\)) increases, based on the derived equation \(v \propto L^{-1}\). Larger animals, despite having more muscle mass and thus greater total power, move slower due to their larger length impacting speed efficiency.

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Key Concepts

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

Power and Work Energy
In physics, power is defined as the rate at which work is done. It tells us how quickly energy is being transformed or transferred. For animals, generating power involves performing actions like moving limbs and overcoming gravitational forces while running up a hill.

Work is the energy transferred by a force acting over a distance, and it can be calculated by multiplying force and distance, both in the direction of the force. When an animal moves up an incline, it expends energy to lift its body against gravity, thereby doing work equal to its gravitational potential energy increase.

To determine the maximum achievable speed of an animal running uphill, we use the concept of the animal's power output. If the power output is proportional to the surface area, then the maximum power is expressed as a function of a characteristic length, like height or diameter, squared: \(P_{\max} \propto L^{2}\). This relationship helps us explore how physical dimensions affect an animal's ability to do work at a certain height.
Gravitational Potential Energy
Gravitational potential energy (GPE) is a form of energy that is associated with the position of an object in a gravitational field, commonly represented by the equation \(GPE = mgh\). Here, \(m\) is mass, \(g\) is the acceleration due to gravity, and \(h\) is the height above a reference level.
  • As the animal climbs uphill, its height \(h\) increases, therefore increasing \(GPE\).
  • This energy transformation tells how much gravitational work needs to be done by the animal to elevate itself.
  • The rate of change of this energy, or the power related to GPE, is crucial in linking power to speed.
By equating the GPE with maximum power, we can relate the inclined movement of the animal to its speed. The change in \(GPE\) manifests in the expression for maximum power as \(mgv\), where speed \(v\) is crucial to understand how quickly energy is being used.
Proportional Relationships in Physics
Proportional relationships are foundational in physics as they describe how one quantity changes with respect to another. Understanding these relationships is key in the physics of animal locomotion.
  • In our problem, we see that \(P_{\max} \propto L^{2}\) indicates a direct proportionality between maximum power and the square of the animal’s size-related characteristic length \(L\).
  • Similarly, expressing mass as \(m = \rho L^3\) shows how volume and length are related.
  • Finally, combining these relationships reveals that the speed \(v\) is inversely proportional to the length \(L\), or \(v \propto L^{-1}\), meaning larger animals tend to move slower due to increased length, even if they can exert more power overall.
This inversely proportional relationship helps us understand real-world observations, such as why a small rabbit may scurry uphill faster compared to an elephant, despite the latter’s higher power potential due to its size.

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Most popular questions from this chapter

How many grams of carbohydrate does a person of mass 74 kg need to metabolize to climb five flights of stairs \((15 \mathrm{m}\) height increase)? Each gram of carbohydrate provides 17.6 kJ of energy. Assume \(10.0 \%\) efficiency-that is, \(10.0 \%\) of the available chemical energy in the carbohydrate is converted to mechanical energy. What happens to the other \(90 \%\) of the energy?

A spring gun \((k=28 \mathrm{N} / \mathrm{m})\) is used to shoot a \(56-\mathrm{g}\) ball horizontally. Initially the spring is compressed by \(18 \mathrm{cm} .\) The ball loses contact with the spring and leaves the gun when the spring is still compressed by \(12 \mathrm{cm} .\) What is the speed of the ball when it hits the ground, \(1.4 \mathrm{m}\) below the spring gun?

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A skier starts from rest at the top of a frictionless slope of ice in the shape of a hemispherical dome with radius \(R\) and slides down the slope. At a certain height \(h,\) the normal force becomes zero and the skier leaves the surface of the ice. What is \(h\) in terms of \(R ?\)

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