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Posted by on Tuesday, November 1, 2011 at 8:00pm.

A force of -9000 N is used to stop a 1500kg car traveling at 20 m/s. What braking distance is needed to bring the car to a halt?

  • physics - , Thursday, November 3, 2011 at 10:51pm

    a = F / m = -9000 / 1500 = -6m/s^2.

    d = (Vf^2 - Vo^2) / 2a,
    d = (0 - (20)^2) / -12 = 33.33m.

  • physics - , Wednesday, May 20, 2015 at 7:52am

    Kinetic energy of car before breaking:
    Wk = mass * velocity^2/2
    Wk = 1500 * 20^2/2
    The work done by the car to transfer kinetic energy to frictional heat:
    W(friction) = Force * Distance
    -9000n * Distance
    As no energy can be lost only transformed, we can write: 1500 * 20^2 /2 = -9000n * distance
    solving for distance we get:
    Distance = 1500 * 20^2/2 * -9000
    Distance = -33,333
    As distance can't be negative and is a scalar quantity we can write 33.333m

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