20 thoughts on “Homework #11”

  1. eotero

    Hi professor!

    So for the homework, for part A, I found the net force to be kinetic friction=uk*n=XXXXX N. Then, I multiplied this force by the distance traveled, since work=fd, which I then got =XXXX J. Is the energy spent=change in Chemical energy which is in this case =work? Because that is how we found the paid energy for Sample Problem 2 in class on Wednesday.

    Thanks for you time.
    Erika

  3. slashley

    For the online hw, I don’t understand problem 11.10 and Cooling a Soft Drink. I know how to convert Kelvin to Celsius but that’s not the right answer… and I don’t know what the hint means by \imagine increasing Tk by 1 degree; by how much would Tc have to change for the temperature equation to remain correct?\ I have no idea how to go about that problem or 11.10 even after having the mass of the rider.
    Can you help, please? Are there any specific equations I should be looking at?

  4. physics131-willocq Post author

    For problem 11.10, here are some steps:
    1. calculate the chemical energy corresponding to the amount of gasoline given (using the fact that 1 gallon corresponds to 3.2 kg of gasoline and 1 gr of gasoline corresponds to an energy of 44 kJ),
    2. calculate the amount of time Dt this amount of energy will last if it is burned at the rate of 480 Watts,
    3. calculate the distance traveled during time Dt at speed 15 km/hr.

    This is similar to the bird problem from online homework #10.

  5. physics131-willocq Post author

    For the temperature difference, try working out an example in which you have two temperatures, say T_1 = 150 K and T_2 = 100 K. The temperature difference T_1 – T_2 is 150 – 100 = 50 K.

    Do the same thing by first converting temperatures T_1 and T_2 to degree Celsius, and then compute the temperature difference between those expressed in degree Celsius. What do you find?

  6. rdetroy

    So for question 11.10, do we find the change in chemical energy by converting 3.2 grams to J’s using the fact that 1 gram of fat=9.4 calories, and that 1 calorie =4186 J ? When I did this conversion and then found the change in time using change in E thermal= 480 W (change in time), I am still getting a wrong answer. Just let me know if you have any suggestions.

  7. slashley

    For 11.10.. I tried your steps and tried doing it the same way as the bird problem. Did 480W = (3.2 x 2.8) over t. I got the answer for time and multiplied this by the 15km. Still not getting the answer. What am I doing wrong?

  8. slashley

    and where in these equations is the person’s mass supposed to come into play? Confused.

  9. physics131-willocq Post author

    Did you take the number of gallons into account? One gallon corresponds to 3.2 grams but the problem gives you a different number of gallons. Also, here we have 1 gr of gas giving 44 kJ of energy (see the info I gave at the start of the last lecture).

  10. slashley

    Yes I took that into account. I only have one more chance at trying to answer and I still don’t know how to do it.

  11. physics131-willocq Post author

    The 68 kg of the person comes into play only indirectly. The input necessary for this problem (which I gave in the announcements on Friday) all comes from Table 11.4 on p.339 of the textbook. It gives the power expended by a person of mass 68 kg to ride a bicycle at 15 km/hr to be 480 Watts. (A person will larger mass would expend more energy to maintain that same speed.)

  12. epickhol

    the problem says that a gallon of gas corresponds to 3.2 kg as opposed to grams. I think this is where people are going wrong maybe. So convert to grams and then try.

  14. dfrancis

    HI Professor,

    I am trying to figure out 11.10 of the online homeowrk. I got that 1,728,000 Joules are needed to power a bicycle at 15 km/hr. The energy that can be used from the gas is 114,400 J. Since we only have a fraction of the energy needed to power the bike for 15km, I tried dividing 114,000 J over 1,728,000 and multiplying that by 15 km. Can you tell me where I am going wrong? Thanks.

    David

  16. jkern

    Also, for question 11.58 I can’t seem to figure out where to even begin. A few things I tried were converting the MW to J and then using these in the efficiency equation with no luck.

    On 11.59 I tried the equation for the maximum efficiency of a heat engine with no luck as well. Am I just missing something on these three or am I on the right track but just mixing things up?

  17. physics131-willocq Post author

    For problem 11.5, you need to calculate the energy required to raise the 15 kg object by a certain height (i.e. the increase in gravitational potential energy).

    For problem 11.58, you just need to consider, say, a 1 second time interval during which a certain amount of energy in Joules is expended (energy per unit time is the power given in MegaWatts).

    For problem 11.59, make sure to convert the temperatures to Kelvin.

  18. physics131-willocq Post author

    For 11.10 David, I don’t understand how you calculated the energy needed to power a bicycle since you do not know the time interval. You need to first calculate the energy available in the gasoline and then figure out how long this energy supply lasts when it is expended at the rate of 480 Watts.

  19. nmetayer

    Professor,
    I’m still having trouble completing the last two problems on thermal efficiency and it’s driving me crazy. Although I did get part B to 11.58 easily, I am completely unsure on how to apply the temperature difference to the problem to get part A and 11.59

  20. physics131-willocq Post author

    For problem 11.58 part A, you need to calculate the maximum efficiency using the cold and hot reservoir temperatures. Here you need to make sure to convert the temperatures to Kelvin. The same applies to problem 11.59.

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