Because we live on the two dimensional surface of a three dimensional world, the tools we have developed in the last two modules are exactly what we need to solve many important problems. In this section we discuss several such problems. You will do most of the work in the exercises, applying geometry to answer these questions.
Work and energy
Work is one of the most difficult physical concepts to understand. Although we all have a good deal of first hand experience with work, that experience can be misleading. Consider, for example, the following tasks. Based on your intuition and experience, compare the amount of work required for each task.
Even though it seems "harder" to lift a thousand pound weight than a hundred pound weight, it takes exactly the same amount of work to lift a thousand pound weight 0.2 feet as a hundred pound weight two feet. In fact, all of the tasks in the first group of four require exactly the same amount of work. One way to see, for example, that lifting a hundred pound weight two feet requires the same amount of work as lifting a fifty pound weight four feet is to imagine cutting the hundred pound weight into two equal pieces -- each weighing fifty pounds. Then lift each piece two feet. This requires exactly the same amount of work as lifting one fifty pound weight four feet and the net result is to lift the original hundred pound weight two feet.
Each of the tasks in the second group of three involves the same work -- the only difference is how rapidly the work is done.
Work and energy are two sides of the same coin. Energy is the capacity to do work. It is often useful to track the energy in a system. Consider, for example, what happens when an object falls from a height of h meters to the ground under the influence of gravity. The work done by a constant force F as an object moves a distance D is
When gravity acts on an object whose mass is m kilograms, the force is
When this object falls a distance of h meters the work done by gravity is
This work is stored in the form of kinetic energy. Note that a joule is the amount of work done when a force of one newton acts on an object as it moves a distance of one meter.
2
mv
Kinetic energy = ----
2
Using the fact that the work done by gravity as an object of mass m falls a distance of h meters is given by 9.80mh joules, determine how fast an object falling from a height of h meters is moving just before it hits the ground. answer
y'' = -9.80, y(0) = h, y'(0) = 0.
Find the function y(t) that is described by the initial value problem above.
Next find out when the object hits the ground by solving the equation
y(t) = 0
Then find out how fast it is going at the instant just before it hits the ground.
Finally, using this information compute kinetic energy as a function of velocity. answer
It is important to keep track of sign and direction when we work with force. For example, when an object drops from an height of h meters to the ground the difference between its ending position and its starting position is -h meters and the force exerted by gravity is -9.80 m newtons. Thus, the amount of work done by gravity is
In two and three dimensions the work done by a force acting on a moving object depends on the strength of the force, the direction of the force and the direction of the motion. The picture below illustrates the force exerted by gravity. This force is pointing downward and can be described by the vector
for an object of mass m.
The movie below shows three objects moving. One object is moving from left to right, staying at a constant height. Because this object's height is constant, gravity does no work as this object moves. A second object (green) is moving from the bottom of the screen to the top. Because this object is rising, gravity is doing negative work. The final (magenta) object is moving diagonally from the lower left to the upper right of the screen. Its motion has two components -- a vertical component and a horizontal component. The only component that involves work is the vertical component -- that is, the component that is parallel to the force of gravity.
Motion and Gravity The figure below shows a force, represented by a vector F, acting on an object as it moves. The motion is represented by the vector S.
Since the component of the force in the direction of the motion is
and the length of the motion is ||S||, the work done by the force is
In the preceding module we saw that the triple product
can be used to compute the volume of a parallelepiped. The volume of a parallelepiped is the area of its base multiplied by the component of the third side that is perpendicular to the base. See the left side of the figure below. If we change one label on this figure we get exactly the situation when a wind is blowing on a parallelogram. See the right side of the figure below. On this side the vector w represents the pressure exerted by a wind blowing on the parallelogram and is measured in units of force per unit of area -- for example, newtons per square meter.
The force felt by the parallelogram is the area of the parallelogram multiplied by the component of the wind that is perpendicular to the parallelogram and is computed using the triple product.
With another change of letter the same picture
describes light falling on a solar collector that is a parallelogram. In this case the vector L representing light might be measaured in watts (or joules per second) per square meter. The energy collected by this solar collector is the area of the solar collector multiplied by the component of the sunlight that is perpendicular to the collector and, once again, is computed by the triple product. The result might be measured in units of watts or joules per second.
The first three problems below involve a wind described by the vector
u = (1, 4, 2)
v = (3, 1, 5)
a = (1, 2, 0)
b = (1, 2, 2)
c = (5, -2, 0)
d = (5, -2, 2)
a = (3, 2, 5)
b = (6, 3, 2)
c = (4, -4, 4)
The problems below involve a solar panel whose four vertices are at the points
a = (0, 0, 0)
b = (0, 5, 1)
c = (10, 0, 0)
d = (10, 5, 1)
Notice that the north edge of this panel is raised slightly, so that it is facing toward the south. This is appropriate for a solar collector in the northern hemisphere. Suppose that if the sun were directly overhead its intensity (measured by a horizontal collector) would be R watts per square meter.