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Homework Help: Net force problem

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data

    An advertisement claims that a particular automobile can "stop on a dime". What net force would actually be necessary to stop an automobile of mass 850 kg traveling initially at a speed of 51.0 km/h in a distance equal to the diameter of a dime, which is 1.8 cm? Answer must be in two sig. figs.

    2. Relevant equations

    V² = u² + 2 a*s
    F(net) = m*a

    3. The attempt at a solution

    51 km/h = 14.16 m/s

    0= (14.16)^2 + 2a(0.018)
    a= 200.5 / -0.036
    a= -5,569.4 m/s^2

    850 kg * -5569.4 = Fnet

    Fnet = 4,734,027.78 N
    (this is 2 sig figs right)?

    Did I do this right? If I did, why exactly do I receive -0.036?(I just multiply 0.018 but I don't really understand why.)

  2. jcsd
  3. Oct 1, 2011 #2
    Your answer has 2 decimal places. Two sig figs would be [tex]4.7 * 10^6 N[/tex]. That's probably what they're looking for. You dont' get the negative value from multiplying, that came from solving your equation for "a."

    [tex]V_o^2=-2a_x\Delta x[/tex]
  4. Oct 1, 2011 #3
    Oh ok thanks. I forgot about that notation.

    Ok, even without the negative value, the final answer should still be the same right?(positive)
    Last edited: Oct 1, 2011
  5. Oct 1, 2011 #4
    No, it should be negative because if you're taking the direction of the initial velocity to be in the +x direction (which you are of course) the acceleration is in the opposite (negative) direction because it is bringing the car to a stop. For the force, you wrote:

    850 kg * -5569.4 = Fnet

    This is correct, but you have to keep the negative when you multiply. You're multiplying a negative and a postive, the result is [tex]-4.7*10^6 N[/tex]
  6. Oct 1, 2011 #5
    Thanks for the clarification. Much appreciated.
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