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Finding the Force physics

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data
    M= 1320 kg in the (+)x-direction
    V= 21.5 m/s
    Find Force?


    2. Relevant equations

    F=MA

    3. The attempt at a solution

    (1320kg)(21.5 m/s) = 28380?
     
  2. jcsd
  3. Oct 21, 2012 #2
    1. Mass cannot have a direction. Did you mean the velocity?
    2. You cannot solve the equation with the information you provided. Please use the homework help template provided to you. If this is from a textbook or homework assignement, copy the question exactly as it is given to you.
     
  4. Oct 21, 2012 #3
    The question:

    A 1320-kg car moves in the + x-direction with speed 21.5 m/s. Assuming constant braking and drag forces, find (a) the force and (b) the work needed to stop the car in a distance of 145 m.
     
  5. Oct 21, 2012 #4
    Much better. Now we can solve it.
    Before we start, though, you're supposed to take drag into consideration? Something seems wrong with that. Please write out the question exactly as it was given in your book or homework assignment.
     
  6. Oct 21, 2012 #5
    That is exactly the question stated in the book :/. That's where I'm confused about too as well.
     
  7. Oct 21, 2012 #6
    OK. We'll just go with the assumption that all the only forces acting on the care are gravity, the normal force of the road and a singular, constant force in the direction opposite the car's motion. You don't need to, but it would be good practice to draw a free body diagram of the car. You can check with the diagram that the only force we need to worry about is that constant force in the negative x direction.

    You are given a constant force, and a constant mass, so there should be a constant acceleration. Try it from there.
     
  8. Oct 21, 2012 #7
    I think the statement "Assuming constant braking and drag forces" means the rate of (-)acceleration and drag force are constant throughout the problem ie no impulse.
    What are the initial and final velocities in the problem?
     
  9. Oct 21, 2012 #8
    Impulse is the change of momentum. Since the car goes from 21.5 m/s to 0 m/s, there is a change in momentum, and therefore an impulse.

    This is a badly worded problem, but I'm almost certain it's telling you that there is a single constant force, therefore constant acceleration. What are the equations of motion for constant acceleration?
     
  10. Oct 21, 2012 #9
    I don't see a constant force being applied to the car? I drew the free body diagram and I did get the assumption there is no value in the x direction, hence that's what I'm looking for, but I honestly a bit stumped.
     
  11. Oct 21, 2012 #10
    What forces do you have acting on the car?
     
  12. Oct 21, 2012 #11
    -9.8N going up, 9.8 down, and 21.5 m/s in the + x-direction.
     
  13. Oct 21, 2012 #12
    What are the initial and final velocities? What is a change in velocity called?
     
  14. Oct 21, 2012 #13
    Initial would be 0, and final would be 21.5 m/s. Change of velocity is the acceleration
     
  15. Oct 21, 2012 #14
    Would the initial velocity be 0 or 21.5? Is there some equation that relates initial velocity, final velocity and acceleration?
     
  16. Oct 21, 2012 #15
    The 21.5m/s is a velocity, not a force. The velocity changes, so there has to be a force somewhere. The problem specifies a force F in the negative x direction. This gives you a constant a=F/m, also in the negative direction. What are the equations of motion with a constant acceleration?
     
  17. Oct 21, 2012 #16
    V=v0+Δat
     
  18. Oct 21, 2012 #17
    Is there an equation relating v, v0, a, distance?
     
  19. Oct 21, 2012 #18
    So I would use V2=V20+2a(X-X0)?
     
  20. Oct 21, 2012 #19
    There should be one that relates distance, initial velocity, final velocity, and acceleration, without time. If you don't know it, use x=x0+v0*t+1/2*a*t^2 and solve for t.
     
  21. Oct 21, 2012 #20
    You're missing a square on the left hand side.
     
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