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Work-Kinetic Energy Theorem with a Variable ForceEquation 7-32 gives the work done by a variable force on a particle in a one-dimensional situation. Let us now make certain that the work calculated with Eq. 7-32 is indeed equal to the change in kinetic energy of the particle, as the work-kinetic energy theorem states. Consider a particle of mass in which we use Newton's second law to replace
in Eq. 7-37 as From the "chain rule" of calculus, we have
and Eq. 7-38 becomes
Substituting (7-40) into 7-37 yields
Note that when we change the variable from Recognizing the terms on the right side of Eq. 7-41 as kinetic energies allows us to write this equation as which is the work-kinetic energy theorem. Sample Problem 7-9 Force SOLUTION; The Key Idea here is that the force is a variable force because its The positive result means that energy is transferred to the particle by force F. Thus, the kinetic energy of the particle increases, and so must its speed. 7-7 Power A contractor wishes to lift a load of bricks from the sidewalk to the top of a building by means of a winch. We can now calculate how much work the force applied by the winch must do on the load to make the lift. The contractor, however, is much more interested in the rate at which that work is done. Will the job take 5 minutes (acceptable) or a week (unacceptable)? The time rate at which work is done by a force is said to be the power due to the force. If an amount of work
The instantaneous power
Suppose we know the work The SI unit of power is the joule per second. This unit is used so often that it has a special name, the watt (W), after James Watt, who greatly improved the rate at which steam engines could do work. In the British system, the unit of power is the foot-pound per second. Often the horsepower is used. Some relations among these units are And We can also express the rate at which a force does work on a particle (or particle like object) in terms of that force and the particle's velocity. For a particle that i! moving along a straight line (say, the x axis) and is acted on by a constant force / directed at some angle
or
For example, the truck in Fig. 7-13 exerts a force
•CHECKPOINT 5: A block moVes with uniform circular motion because a cord tied to the block is anchored at the center of a circle. Is the power due to the force on the block from the cord positive, negative, or zero? Sample Problem 7-10 Figure 7-14 shows constant forces (a) What is the power due to each force acting on the box at that instant, and what is the net power? Is the net power changing at that instant? SOLUTION: A Key Idea here is that we want an instantaneous power, not an average power over a time period. Also, we know the particle's velocity (rather than the work done on it). Therefore, we use Eq. 7-47 for each force. For force
Fig. 7-14 Sample Problem 7-10. Two forces Fx and F2 act on a box that slides rightward across a frictionless floor. The velocity of the box is v*.
Date: 2015-01-12; view: 1809
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