Power

Practice

practice problem 1

A typical adult in the United States consumes something like 2000 dietetic calories of food per day. Determine the average power generated by such an adult (assuming he or she is not gaining or losing weight).

solution

We have to assume here that all the food energy consumed goes into work on some level (mechanical or metabolic). This is true only so long as the person is not gaining or losing weight, which occurs when the energy consumed is greater than or less than the work done. Start the problem by converting the units — convert the energy from calories to joules and convert the time from days to seconds.

P =  W  =  2000 kcal   4186 J   1 day  = 97 W
t 1 day kcal 24 × 60 × 60 s

With something between 700 and 800 watts in a horsepower, this corresponds to a rate of energy conversion somewhere between one-seventh and one-eighth of a horsepower.

practice problem 2

Determine the cost of operating a 7000 Btu, room-sized air conditioner in New York City for the duration of the summer. Assume that electricity costs 14¢ per kilowatt hour and that the air conditioner will run about 10 hours a day for 80 days.

solution

In the United States, the Btu is often confused with the Btu per hour (Btu/h). This is partially the fault of the heating, ventilation, and air conditioning industries (HVAC). The rate at which energy is transformed from one form to another or transferred from one place to another is called power. Power is stated in units of energy divided by time. Furnaces and air conditioners are rated according to their heating and cooling power; that is, how quickly they can add or subtract heat from a room or home. American appliance manufacturers often quote the power of their devices in "Btu" when they really mean "Btu/h". Most certainly, this is just a form of shorthand and does not reveal any malicious intent on the part of the industry. However …

Electric energy is sold by the kilowatt hour while air conditioners are rated in Btu/h. This makes it extremely difficult to estimate the operating costs of these energy hungry appliances. For example, a 7000 Btu/h room air conditioner consumes energy at a rate of about …

7000 Btu   1055 J   1 h  = 2050 W = 2.1 kW
1 h 1 Btu 3600 s

while electricity in the New York City metropolitan area averages about 14 ¢ per kilowatt hour. Thus, every ten hours of use costs …

2.1 kW   10 h   14 ¢  = 290 ¢ = $2.90
1 1 kWh

Given that summer days in NYC range from semitropical to subtropical to absolutely tropical, it's quite reasonable to assume 80 days of air conditioner use in a typical season. This brings the cost of cooling one room to …

$2.90   80 AC days  = $230/AC season
AC day AC season

This cost may or may not be acceptable to an individual consumer for this particular use. That isn't the point. The point is that rating an air conditioner in a nonstandard unit adds one more step to the problem. Appliances should be rated in watts or kilowatts so that consumers would be able to make mental estimates of their operating costs more easily.

practice problem 3

Write something calculus based if possible.

solution

practice problem 4

The athlete in this video clip [mov] is performing a weightlifting maneuver known as the snatch. In this maneuver, the barbell must be lifted from the platform to a point above the head, with the arms and legs fully extended, in a single movement. The barbell must then be held motionless until the referees give the signal and then returned to the platform. In this particular video …
  • the mass of the barbell is 77.5 kg (for comparison, the mass of athlete is 58 kg),
  • the disks on the barbell have a diameter of 450 mm, and
  • the video advances at 25 frames per second (with 71 frames total).
Determine the following quantities for the barbell this athlete is lifting as functions of time …
  1. height
  2. velocity
  3. acceleration
  4. applied force
  5. work
  6. power
Most of this question is a review of mechanical concepts discussed in previous sections in this book. Only the last part deals with power. To begin this problem, you will need some sort of screen measuring tool. Many basic image editing applications have this function built into them. If your doesn't, you might like to try Wadruler [exe] for Windows or Freeruler [external link] for Mac OS X.

solution

  1. Height
    Determine the position of some easily identifiable point on the barbell in any units that are convenient. For calibration purposes, measure the diameter of the disk on the end of the barbell in the same units. Set up a proportion to convert the position measurements in arbitrary units to height in meters. Set up a similar proportion to convert the frame number to elapsed time in seconds.
                 
    h =  height on screen  0.450 m   t =  frame number  ⇒ 
    diameter on screen 25
 
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  1. Velocity
    By definition, velocity is the rate of change of position with time. Take the slope of the line tangent to the height-time graph to produce the velocity-time graph or, in the language of calculus, take the derivative of height with respect to time.
         
    v =  dh  ⇒ 
    dt
 
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  1. Acceleration
    By definition, acceleration is the rate of change of velocity with time. Take the slope of the line tangent to the velocity-time graph to produce the acceleration-time graph or, in the language of calculus, take the derivative of velocity with respect to time.
         
    a =  dv  ⇒ 
    dt
 
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  1. Applied Force
    A net force causes an acceleration. The net force in this situation is a combination of the force applied by the athlete and the weight of the barbell. Since these forces point in opposite directions, their vector sum is difference of their magnitudes. Thus, the applied force of the athlete (the force we care about) is the net force on the barbell plus its weight.
         
    ∑  F = ma = F − W = F − mg  ⇒ 
      F = m(a + g)
 
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  1. Work (Three Steps)
   
    1. Change the horizontal axis from time to height. Since the athlete lowered the barbell for a brief moment about halfway into the lift, the graph is a bit twisted in the middle.
 
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    1. Work is defined as the cumulative product of force and displacement. Use the area under the force-height graph to produce the work-height graph. In the language of calculus, the work is the force-displacement integral. Like the force-height graph, this graph is also bent out of shape.
         
      W = ∫ F dh  ⇒ 
 
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    1. Change the horizontal axis back from height to time. Since time only marches forward, the graph is a classic, single-valued function again. Work can only have one value at any instant in time.
 
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  1. Power
    By definition, power is the rate at which work is done. Take the derivative of the work-time graph to get the power-time graph. Weightlifting is an exceptionally powerful sport. This athlete had an average power of 390 W and a maximum power of 1900 W (approximately ½ and 2½ horsepower, respectively).
         
    P =  dW  ⇒ 
    dt
 
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