Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Kinetic Energy Vs. Potential Energy confusion.

  1. Oct 25, 2007 #1
    1. The problem statement, all variables and given/known data

    A car and driver weighing 5530 N passes a
    sign stating "Bridge Out 25.5 m Ahead." She
    slams on the brakes, and the car decelerates
    at a constant rate of 13.7 m/s^2 :
    The acceleration of gravity is 9.8 m/s^2 :
    What is the magnitude of the work done
    stopping the car if the car just stops in time
    to avoid diving into the water? Answer in
    units of J.

    2. Relevant equations
    This is where I'm stuck. In class today we learned about Conservation of Energy and I don't know if that applies here.
    I know there is constant acceleration in forward direction so I could use Kinematics but I don't know where to go from there I'm confused as to what sign to designate to 13.7 m/s^2.
    So far I have: Vf^2=Vo^2+2a(deltaX)
    0=Vo^2+2(-13.7 m/s^2)(25.5m)
    Vo=26.432934 m/s

    3. The attempt at a solution
    Last edited: Oct 25, 2007
  2. jcsd
  3. Oct 25, 2007 #2
    Anyone out there to help?
  4. Oct 25, 2007 #3


    User Avatar

    If you want to solve it with energy, I suggest beginning with all the pertinent conditions at the beginning and setting them equal to those at the end (think what kinetic will be) plus the energy lost by heat in the breaks, which is what you're looking for.
  5. Oct 25, 2007 #4
    Nevermind I just figure out the entire problem.
  6. Oct 26, 2007 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    using the kinematic equation gets you the initial velocity of the car, which you have correctly done. Now if you want to use conservation of energy,write down that equation and solve for the work done by the braking force. PE does not come into the equation, since there is no PE change. (It's perhaps a bit easier to use Newton 2, but that depends on how you are asked to solve this problem.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook