practice problem 1
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.
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
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…
|P =||7000 Btu||1055 J||1 h|
|1 h||1 Btu||3,600 s|
|P = 2050 W = 2.1 kW|
…while electricity in the New York City metropolitan area averages about 14 ¢ per kilowatt hour. Thus, every ten hours of use costs…
|cost/day =||2.1 kW||10 h||14 ¢|
|cost/day = 290 ¢ = $2.90/AC day|
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…
|cost/season =||$2.90||80 AC days|
|AC day||AC season|
|cost/season = $230/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. All appliances should be rated in watts or kilowatts so that estimating their operating costs can be done more easily.
practice problem 3
- the average stopping force applied to the ship and
- the average power dissipated while stopping it.
- This problem lends itself well to standard techniques. State the given quantities and convert them to SI units as needed.
Δs = 13 km = 13,000 m m = 1.50 × 106 kg v = 0 m/s v0 = 50,000 m = 13.889 m/s 3,600 s
Use the Work-Energy Theorem. The change in the supertanker's kinetic energy is due to the work done by its engines running in reverse. (It can also be solved using Newton's Second Law of motion — as it was an earlier section of this book. Both methods lead to exactly the same answer — as they should.)
W = ΔK FΔs = 1 mv02 2 F = mv02 2Δs F = (150 × 106 kg)(13.889 m/s)2 2(13,000 m) F = 1,112,900… N = 11 MN
- Let's try this part two ways and see if different solutions give us the same answer.
Method 1: Power is the rate at which work is done. The work done is equal to the loss of kinetic energy.
W = ΔK W = 1 mv02 2 W = 1 (150 × 106 kg)(13.889 m/s)2 2 W = 14,567,659,259 J = 14.5 GJ
The time it takes to stop can be found using the two formulas for average velocity.
Δs = v + v0 Δt 2 Δt = 2Δs v + v0 Δt = 2(13,000 m) 13.889 m/s Δt = 1,872 s = 31:12
Combine work and time to get power.
P = W t P = 145,6765,926 J 1,872 s P = 7,728,415 W = 7.7 MW
Method 2: Power is also the product of force and velocity; or to be more precise, the product of average force and average velocity. We calculated the average force in part a. The average velocity is the average of the initial and final velocities. No surprise here. We get the same answer as before, but with much less effort.
P̅ = F̅v P̅ = 1,112,900 N ⎛
13.889 m/s + 0 m/s ⎞
2 P̅ = 7,728,415 W = 7.7 MW
The first sentence of this question came from an article in the Chicago Tribune dated 6 April 2001, "Strait with No Equal Has New Danger." Moral of the story: Yield to the supertanker when you're sailing the Bosphorus.
practice problem 4
- 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).
- applied force
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 diameter on screen t = frame number 1 s 25
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
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
- Applied Force
A net force (∑F) causes an acceleration (a). The net force in this situation is a combination of the force applied by the athlete (F) and the weight of the barbell (W). 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 = ma F − mg = ma F = m(a + g)
- Work (Three Steps)
- 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.
- 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 = ⌠
- 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.
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