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Motion Near Earth's Surface

  1. Jan 5, 2007 #1
    1. The problem statement, all variables and given/known data

    Hi, I just want to know if I am on the right track with the answers I have came up. Thanks for everyone's help.

    1. (a) Cathy takes the bus home form work. In her hand she holds a 2.0 kg cake box, tied together with a string. As she ascends the steps into the bus, the box accelerates upward at a rate of 2.5 m/s^2. What is the force exerted on the string?

    (b) Cathy sets the box on the seat beside her. The bus accelerates form rest to 60.0 km/h in 4.0 s, and the box begins to slide. What is the coefficient of static friction between the box and the seat of the bus?

    (c) A taxi suddenly cuts in front of the bus, causing the bus driver to slam on the brakes. The bus driver reduces speed from 60.0 km/h to 20.0 km/h in 1.5 s. Does Cathy's cake slide forward?

    2. Relevant equations


    formula for finding the coefficient of friction (I think....)

    U = F(k) / F(n)

    3. The attempt at a solution

    (a) F= m(a)
    =(2.0 kg) (2.5 m/s)
    =5 N

    (b) U = F(k) /F(n)
    =16.67 m/s /5 N
    = 3.3

    (c) My first inclination would be that Cathy's cake would slide. However, in tring to figure out how to prove this I wasn't quite sure except posssibly using the distance formula.
    d=1/2 (v1 + v2) t
    =1/2(60 m/s + 20 m/s) 1.5
    =60 m
    (this seems like an awful big number and doesn't make sense)

    Am I missing a certain concept here?
    Thanks for the help
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 5, 2007 #2
    You should draw a free body diagram (FBD) to understand this better. You have the force of the string, and the force of gravity, and some kind of motion. You need to write the equation for the net force and solve that.

    I don't even have to check the math to know this is not right. Look at your units. Remember that Force has units of Newtons. You are dividing a velocity by a force.

    I would suggest solving most of these problems symbolically, and only adding the numbers as the final step. It helps reduce errors, and it can provide a bit more insight.

    You should again draw a FBD to see the forces more clearly. Write the equation, and see what you get.

    Last edited: Jan 6, 2007
  4. Jan 5, 2007 #3
    okay part A is good.
    for part B, first you should know that it is wrong because your U should be between 0 and 1. Now I think you should make sure you are converting to m/s(sq) correctly. Make sure you first change to killometers per hour and then divide by 60 then by 60 again. Multiply your acceleration by the mass to find the frictional force. I think you are good from there.
    For part C, first do the same thing as in part B but you can once you find the force on the block, you just need to see if it is more or less than the frictional force that you already calculated.
  5. Jan 6, 2007 #4
    I don't agree. You are both neglecting the contribution of gravity.
  6. Jan 6, 2007 #5
    I thought the equation for net force was F(net) =m(a)
    I agree that you should take into account gravity (9.8 m/s^2) but how would you incorporate that into the above equation?

    :confused: Pharm 89
  7. Jan 6, 2007 #6
    The forces on the box replace F_net.

    I really would encourage you to draw FBD diagrams, even for these simple problems. It's a good habit to develop.

  8. Jan 6, 2007 #7
    i substituted 9.8 m/s^2 for F_net, still using the same varibles for mass and acceleration and solved for X. and Icame up with 0.51 N?????

    Or is my method simply wrong.
  9. Jan 6, 2007 #8
    9.8 m/s^2 is an acceleration. F=ma, so you have to multiply it by m to get a force. I hope you don't mind me mentioning that you make this mistake a lot, maybe just something to remember to check.

    On your FBD diagram, you'll have two arrows. One pointing up, that represents the force of the string. One pointing down, that represents the force of gravity (these are vector forces). You can put a third arrow off to the side, pointing up, to indicate the direction of acceleration, that's helpful too.

    Then select a direction to consider positive, let's say up is positive. The vector sum of these two forces is the net force, and equals ma. That's the meaning of the second law.

  10. Jan 6, 2007 #9
    Thanks for your help with this problem. It makes more sense to me know.
  11. Jan 6, 2007 #10
    You're welcome, it's my pleasure.

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