1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Friction (3 part question)

  1. Jul 12, 2009 #1
    1.A grocery cart has a mass of 32.0 kg. An applied force of 4.00 x 10^2 N [E] is used to move the cart. The cart starts from rest and the force is applied for 5.0s.

    a) Calculate the force of friction acting on the cart if the coefficient of friction between the cart and asphalt is .87
    b) Calculate the acceleration of the grocery cart
    c) How far does the cart move in 5.0s?


    Edit: since there is no vertical movement Fnety = 0 and Fg = mg
    therefore Fn = mg
    Fn = (32 kg)(9.81m/s^2)
    = 313.92 N
    =310 N (sigdigs)

    Ff = mu(Fn)
    =.87 x 310
    = 269.7 N
    = 270 N [W]

    but i have no idea if thats accurate or not =\ i started on part b, which is dependant on part a so if i dont have this correct then i cant continue

    Edit two
    i began part b ....
    so...
    Fnetx = (4.0 x 10^2 N [E]) - 270 N [W]
    = 130 N

    F = ma
    130 N = 3.2 kg (a)
    a = 4.06 m/s^2
    a = 4.0 m/s^2
     
    Last edited: Jul 12, 2009
  2. jcsd
  3. Jul 12, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Welcome to PF.

    I see the three part question, but I don't get much of a sense of 3 part effort on your part to address the problem.

    Care to share what you are having a problem with?
     
  4. Jul 12, 2009 #3
    Oh sorry >.< Okay so i tried finding the friction
    since there is no vertical movement Fnety = 0 and Fg = mg
    therefore Fn = mg
    Fn = (32 kg)(9.81m/s^2)
    = 313.92 N
    =310 N (sigdigs)

    Ff = mu(Fn)
    =.87 x 310
    = 269.7 N
    = 270 N [W]

    but i have no idea if thats accurate or not =\ i started on part b, which is dependant on part a so if i dont have this correct then i cant continue
     
  5. Jul 12, 2009 #4

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Rounding 313.92 N to three sig. dig. precision gets you 314 N, NOT 310 N. Can you see that 310 N would not make any sense? If some force was exactly 313.92 N, but you had a weigh scale to measure it that was only accurate to the nearest newton (i.e. the markings on the dials on the weigh scale are in 1 N increments), then your measurement of this force will cause the needle to appear very close to the 314 N marking. You would conclude that (to the accuracy of your measuring device), the force was 314 N.

    In this case you aren't doing the measurements yourself, but you are assuming that the numbers quoted in the problem are measured quantities, not exact ones. Therefore, you are limiting your calculated results to have the same precision as that of the least precise given measurement, therefore using it to determine the appropriate number of sig digs for the answer (a number that ensures you are not overstating the precision of your result and therefore overstating the certainty with which you know it).

    If you intended to express it to two sig digs, then remember that trailing zeros count as sig digs. Therefore, you cannot state it as 310 N. You would have to say 3.1 x 102 N. Because your precision is limited to two sig digs, all you can say is that the measurement is somewhere between 310 and 320 N, but you don't know it any more precisely than that, because you can't measure that third digit.

    Anyway, your calculation of the normal force looks fine, aside from that error. Your equation for the frictional force also looks correct (although you must correct that error in order to get the right friction force).
     
    Last edited: Jul 12, 2009
  6. Jul 12, 2009 #5
    Oh okay thank you but then wouldnt i have to state the answer for part a as 3.1 x 10^2 because i cannot state 3.14?


    i began part b ....
    so...
    Fnetx = (4.0 x 10^2 N [E]) - 270 N [W]
    = 130 N

    F = ma
    130 N = 3.2 kg (a)
    a = 4.06 m/s^2
    a = 4.0 m/s^2
     
  7. Jul 12, 2009 #6

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well...32.0 kg and 9.81 m/s2 are both given to three significant digits. So why are you limiting yourself to two for the final answer?


    I haven't checked your arithmetic. Your method looks okay. Okay, so now you know that the cart is moving with some constant acceleration, and you know what that acceleration is. You also know how long it accelerates for. Can you think of some equation that can give you the final velocity of an object that has been accelerating at a constant rate for a certain amount of time?
     
  8. Jul 12, 2009 #7
    We were taught that for sigdigs we use the lowest amount of sigdigs in the given information as the number of sigdigs in our final answer

    So would I find v2 since v1 = 0 ?
    a = v2-v1 over time
    4.o m/s^2 = v2 - 0 over 5s
    20m/s = v2

    therefore subbing in 5 seconds? to find the distance?
    Or can i not sub in 5 seconds twice?
     
  9. Jul 12, 2009 #8

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    LOL. I know that. What I am saying follows exactly from that rule. Since both of the quantities you used to calculate the normal force have three sig digs, the lowest number of sig digs of the two of them is...three. Therefore, the final answer for the normal force should have three sig digs.

    Your work looks correct, although I think you have a fair bit of rounding errors. Although you should express (i.e. write down) your answers to the correct number of sig digs, in multi-step calculations, you should keep the full number of digits for the calculated intermediate values on your calculator, and not round until you get to the end of the multi-step calculation. This avoids rounding errors from being introduced throughout your calculation. So for instance:

    313.92 N * 0.87 = 273.1104 N.

    You would express the answer as 2.7 x 10^2 N, but in the NEXT step, you would keep the full value (273.1104 N), and say:

    Fnet = 400 N - 273.1104 N = 126.8896 N

    a = 126.8896 N / 32.0 kg = 3.9653

    You would express this final answer correctly (to two sig digs due to the friction force) as 4.0, but in the NEXT step you would likewise use the full value of the acceleration (to prevent rounding errors).



    Sorry, I made a mistake here. I thought that the question was, "How fast is the cart moving after 5.0 s?" But it is in fact, "How far does the cart go in 5.0 s?" Therefore, you need to use an entirely different equation, one that gives you the distance travelled given a constant acceleration and a certain time interval.

    No, you can't just sub the 5.0 s in again, because the velocity is not constant.
     
  10. Jul 12, 2009 #9
    Wow thanks for all the help!
    I'm gonna try redoing my questions with your advice =]
    I may be back for another question later tonight haha >.<
    Thanks again~
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Friction (3 part question)
Loading...