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Grade 11 Physics Problem Please Help

  1. Nov 13, 2007 #1
    1. The problem statement, all variables and given/known data
    A Rocket Sled weighing 20000N is gliding at 24 m/s (North) on ice where friction is negligible. Suddenly it passes over a rough patch 22m long, which creates a friction force of 6000N (south). With what velocity does the sled leave the rough patch?


    2. Relevant equations
    Vi=Vf+ad
    Vf(squared)=Vi(squared)+2ad
    Fnet=Ff+Fa
    Fnet= Ma


    3. The attempt at a solution

    I figured that the Vi of the sled when it hits the patch is 24 m/s since there is no acceleration. When it hits the patch, there are two forces on the object; one with -6000N and the other which is the original force applied. The sum of these constitutes the net force. The net force is also equal to the Mass of the object x Acceleration. The mass is 20000/9.81 which equals roughly 2038.7KG.

    If I find acceleration, I would easily find the Vf of the object leaving the rough patch. The problem is, I have no idea how to find the forward force of the rocket sled, so I am left with 2 variables in 1 equation and I am stuck.

    Can someone please tell me the way to solve this? Thank you.
     
  2. jcsd
  3. Nov 13, 2007 #2

    Doc Al

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    Staff: Mentor

    There's only one horizontal force acting on the sled--the friction. There's no "original force" needed to keep the sled moving as it glides over the ice.


    There is no forward force, just friction. Perhaps you are thinking of the force that was needed to get the sled moving at 24 m/s? Once it got moving, no force is needed to keep going. (Newton's 1st law.)
     
  4. Nov 13, 2007 #3
    Hmm what you say makes complete sense. But i still do not reach the correct answer. Using your advice, the modified solution method would be:

    Fnet= -6000N
    Fnet= 2038.7Kg(a)
    a= -2.94.

    Solving this using the Vf(squared)=Vi(squared)+2(a)(d) formula, Vf= 22.6 m/s
    The answer is 23.7 m/s, so I am off by a factor of 1.1. Can you point me to the right direction as to what may have caused this deviation? I'm quite sure it can't be rounding errors.
     
  5. Nov 13, 2007 #4

    Doc Al

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    Staff: Mentor

    Your method is perfectly correct. (I'd say that your answer is a bit too high--recheck your arithmetic!) No reason to think that 23.7 m/s is correct. (Also, realistically speaking, only 2 digits--at most--are significant.)
     
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