The driver of a 810.0 kg car decides to double the speed from 23.6 m/s to 47.2 m/s. what effect would this have on the amount of work required to stop the car, that is, on the kinetic energy of the car? kei= × 105 j kef= × 105 j times as much work must be done to stop the car.

Step-by-step explanation:

The kinetic energy for a given mass and velocity is ...

  KE = (1/2)mv^2 . . . . . m is mass

At its initial speed, the kinetic energy of the car is ...

  KEi = (1/2)(810 kg)(23.6 m/s)^2 ≈ 2.256×10^5 J . . . . . m is meters

At its final speed, the kinetic energy of the car is ...

  KEf = (1/2)(810 kg)(47.2 m/s)^2 ≈ 9.023×10^5 J

The ratio of final to initial kinetic energy is ...

  (9.023×10^5)/(2.256×10^5) = 4

4 times as much work must be done to stop the car.

You know this without computing the kinetic energy. KE is proportional to the square of speed, so when the speed doubles, the KE is multiplied by 2^2 = 4.

according to the triangle sum theorem, all the angles in a triangle must add up to 180 degrees. if we know one angle is already 90, we do 180-90=90. the other 2 angles must add up to 90, exactly.

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