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Homework Help: Basic Mechanics

  1. Jun 13, 2008 #1
    This is a fairly simple problem which i believe i know the answer to but the answer doesn't agree with the book i got it from.

    Theres a 750 gram cart traveling 4 m/s, 2 meters later its traveling at 3 m/s. Calculate the average retarding force assuming only air resistance.

    Here what I did:

    KE= (1/2)*m*v^2 so deltaKE=(.5)*(.750)*(4^2-3^2)= 2.625

    The Work=f*d. Since the Change in Kinetic Energy equals the work done on the object i did:
    2.625=f*2. Hence f=(2.625/2)=1.3125 N.

    I think thats right but is there something i might be possibly missing?
  2. jcsd
  3. Jun 13, 2008 #2
    Looks good to me.
  4. Jun 13, 2008 #3
    I think it is wanting you to calculate the average air resistance (disricbed in force) that it would take to slow a 750 gram object from 4 m/s to 3 m/s in 2 meters.
  5. Jun 14, 2008 #4
    i tried it a slightly different way and got the same answer, only negative. i did

    where u=inital velocity, v=final velocity,a=acceleration and x being displacement.
    so using that i got

    a= (v²-u²)/2Δx


    Fnet= m ((v²-u²)/2Δx)

    plug all the numbers in and you get -1.312N, negative because you are decelerating in the negative x direction.

    the big thing is you must watch your sign. if you have air resistance slowing you its not going to be the same sign as your x direction and velocity. velocity implies direction as well as speed. you can make your velocity negative based on where you allocate your x and y axis relative to your direction, but then your force would be positive. always watch the sign.
    Last edited: Jun 14, 2008
  6. Jun 14, 2008 #5
    "negative because you are decelerating in the negative x direction."

    sorry, that is worded poorly. it should say, becuase you are decelerating in the positive x direction, meaning you are accelerating in the negative direction.
  7. Jun 14, 2008 #6
    Thanks everyone, it has been quite good to see that other people got the same answer as i did...im pretty sure the book just mistyped the answers or something
  8. Jun 14, 2008 #7


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    if it's a recent book it may have a web page with errata on it. If you come across something like this in the future it may be worth looking there first. If its not on the list you could send in the error you've found to the publishers.
  9. Jun 14, 2008 #8


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    I wouldn't have done it that way. At a constant acceleration (i.e. "average" acceleration), a body's "average" speed is "(initial speed+ final speed)/2". Here, the initial speed was 4 m/s and final speed was 3 m/s so its average speed during those 2 meters is (4+3)/2= 7/2 m/s. At that speed, it took 2/(7/2)= 4/7 seconds to travel 2 meters so its speed dropped from 4 m/s to 3 m/s, a difference of -1 m/s, in 4/7 seconds: the (average) acceleration was (-1)/(4/7)= -7/4= -1.75 m/s2. Since the mass is 750 g= .75 kg, the force is (.75)(-1.75)= -1.3125 N as you say.
  10. Jun 15, 2008 #9
    Yeah i couldn't find the errata and it is a fairly new book.
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