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jmays14 Group Title

A 1000 kg car moving at 20. m/s is stopped by the action of its brakes. The work done by the car's brakes is ____ Joules.

  • one year ago
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  1. SilenTempest Group Title
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    K=(1/2)mv^2

    • one year ago
  2. SilenTempest Group Title
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    Plug in the info you have. Post your answer and I will tell yu if its wrong

    • one year ago
  3. JamesJ Group Title
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    To expand slightly on ST's argument: before the breaking, the car had a lot of kinetic energy, KE, given by that formula. At the end it had no kinetic energy. So the work the brakes had to do was the change that KE to zero. As work done on a system is equal to the change of energy of a system, the amount or magnitude of work done by the brakes in this case is equal to the original KE.

    • one year ago
  4. jmays14 Group Title
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    @jamesJ so the answer would be zero

    • one year ago
  5. JamesJ Group Title
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    No. The brakes do work on the car to reduce the KE from (1/2)mv^2 to zero. That is not a zero amount of work. It is -(1/2)mv^2 amount of work.

    • one year ago
  6. JamesJ Group Title
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    So once again: the change of energy, \( \Delta E \) of system of equal to the amount of work, \( W \), done on the system. The energy of the car that is changing is its kinetic energy. It is initially travelling at speed \( v \) and has kinetic energy \( KE_{initial} = \frac{1}{2}mv^2 \), with the values of m and v given. After the brakes have done their work the car's speed is zero hence the kinetic energy of the car is \( KE_{final} = \frac{1}{2}mv^2 = \frac{1}{2}m0^2 = 0 \) Therefore the work done by the breaks is \[ W = \Delta E = KE_{final} - KE_{initial} = 0 - \frac{1}{2}mv^2 = -\frac{1}{2}mv^2 \] Got it?

    • one year ago
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