Question

Coriolis' alternative method for finding work can be used to estimate the relative stopping distance for a car given the car's speed. For this application of Coriolis' alternative method, the force of friction is in the opposite direction of the motion of the car on which that frictional force is acting, and the "work required to stop the car" (sometimes called "the work done by friction") is equivalent to the amount of mechanical energy converted to thermal energy by the action of the car's brakes. The driver of a 780 kg car decides to double the speed from 20.3 m/s to 40.6 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? Give your answers to the following questions in joules.

(a) The kinetic energy at a speed of 20.3 m/s would be

(b) The kinetic energy at a speed of 40.6 m/s would be

(c) How many times more work is required to stop the car now that it is going twice as fast?

Four times as much work is required, therefore (assuming that the amount of friction force when the brakes are applied is a constant) four times as much distance will be required for the car to stop.

The same amount of work is required in each case, therefore (assuming that the amount of friction force when the brakes are applied is a constant) the same amount of distance will be required to stop the car when it is moving at twice the speed.

Two times as much work is required, therefore (assuming that the amount of friction force when the brakes are applied is a constant) two times as much distance will be required for the car to stop.

Three times as much work is required, therefore (assuming that the amount of friction force when the brakes are applied is a constant) three times as much distance will be required for the car to stop.

Answer #1

The driver of a 800.0 kg car decides to double the speed from
20.3 m/s to 40.6 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

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