Physics 12: Homework #2: due April 19


The book has run out of steam on us in regard to quantitative problems on basic forms of energy. This homework aims to give you more practice at calculating the energy content of systems of interest. Some of these are tough problems—harder than exam questions—but they'll give you good practice. Work together and take advantage of office hours to make the most of these.

  1. A watermelon is thrown from an 8 story stariwell (25 meters high) and hits the ground with a splat. How fast is the melon going when it hits the ground (ignoring things like air resistance)? Use the formula for potential energy and assume it all gets converted to kinetic energy. Does it matter how massive the melon is?

  2. A loaded 18-wheeler with a mass of 40,000 kilograms (about 40 cars) moves along the freeway at 30 meters per second (67 mph). How much energy is in the truck's motion? If you want to get the truck up to speed in 20 seconds, how much power does the engine have to provide? If one horespower is 746 Watts, about how many horsepower is this? For reference, a car is usually 100 to 200 horsepower. Note: look at the loose precision of the input numbers (usually only one significant figure). Now look at your answer in horsepower. Did you report more accuracy than is warranted by the rough input numbers? Does it make you feel silly?

  3. When you stick your hand out of the car window, you feel a force from air resistance. You are in effect applying this force through some distance of travel, so you are doing work. You can turn the problem around and ask how much kinetic energy is in the "wind" hitting your hand, and figure out how much power you are supplying to keep your hand there. Assume your hand has a cross-sectional area of 0.01 square meters (e.g., 10 cm square), and that you are traveling at 30 m/s. How much power does it take to hold your hand there, assuming all of the wind's kinetic energy is removed? Air density is 1.3 kg per cubic meter.

  4. The heat capacity of water is 4184 J/kg/°C, meaning that 4184 Joules (1 kcal) will heat 1 kg (1 liter) of water 1 degree Celsius. If you have a microwave oven that puts out 1046 Watts (you'll thank me for this strange number), how long does it take to boil 0.1 liters of water if the water starts at 20 degrees Celsius? Water boils at 100 degrees Celsius. Assume that all of the microwave's power is deposited into the water.

  5. A 10-minute hot shower might use about 80 liters of water, heated to 45 degrees Celsius. If the water starts at 15 degrees Celsius, how much energy has been put in to heat it to 45 °C? Express your answer in Calories (kcal), calories, Joules, and Btu. How does the answer compare to your daily intake of food energy?

  6. If you converted your home to a hot-on-demand system (i.e., no hot water reservoir), how much power would it have to put out to heat water fast enough for the hot shower in the previous problem (has to supply the amount of energy calculated in the last problem in ten minute's time)? Again, watch for excessive, meaningless digits in your reported answer. Compare this to an 1800-Watt blow-dryer (how many blow-dryer's worth is this?).

  7. The kinetic energy of the "wind" hitting the front of a truck moving at 30 m/s with a cross-sectional area of 4 m² is ½mv² = ½(1.3 kg/m³)×(30 m×4 m²)×(30 m/s)2 = 70,200 J in every second (a column of air 30 meters long encounters the truck every second). If all of this kinetic energy is robbed from the air (bad aerodynamics), and you get 10 kcal per gram of gasoline, how many grams per second of gasoline do you need to supply to the truck. Assume the truck's engine is only 25% efficient, so that you need to supply four times as much gas as you might naively calculate.

  8. A gallon of gas contains roughly 3000 grams. Using the value from the previous problem in grams per second needed to supply the truck, how long will it take to run through a gallon? Given that the speed is 30 m/s or 67 miles per hour, how many miles do you get per gallon?

  9. Give the approximate temperature of each of the following in Fahrenheit, Celsius, and Kelvin:
    1. body temperature
    2. room temperature
    3. freezing point of water
    4. boiling point of water
    5. absolute zero (0 °K)
    6. solar surface (5800 °K)
    7. center of the sun (16 million °K)


  10. If I tell you that the intensity of sunlight on the earth's surface is about 1000 Watts per square meter, use this directly to compute the corresponding temperature using F = σT4; σ = 5.67×10-8 W/m²/°K4. This is the temperature (in Kelvin) a perfect (black) absorber would achieve in full sunlight if air convection and conduction to the surroundings didn't carry some energy away. Convert this to °C and °F. Hint: You will have to take a fourth root to get the temperature. On most calculators, use the yx button and then enter 0.25 (for one-quarter power). Try the fourth root of 81, and you should get 3. Alternatively, you can take the square root twice.

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