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jmays14
 2 years ago
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.
jmays14
 2 years ago
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.

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SilenTempest
 2 years ago
Best ResponseYou've already chosen the best response.0Plug in the info you have. Post your answer and I will tell yu if its wrong

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.0To 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.

jmays14
 2 years ago
Best ResponseYou've already chosen the best response.0@jamesJ so the answer would be zero

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.0No. 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.

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.0So 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?
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