If the speed of a car is doubled, the work to stop the car changes by a factor of
so it's 4....but I cannot think of how I know it is double. Any explaination?
It is not double - it is squared,
because KE involves v^2
To understand why the work to stop a car changes by a factor of 4 when its speed is doubled, let's consider the relationship between work, speed, and kinetic energy.
The work done on an object is equal to its change in kinetic energy. When a car is brought to a stop, its kinetic energy is reduced to zero.
Kinetic energy (KE) is given by the formula:
KE = 0.5 * mass * speed^2
Therefore, if the speed of the car is doubled, the kinetic energy will be four times greater.
Now, when the car comes to a stop, all the kinetic energy has to be dissipated to overcome the car's momentum. The work done to stop the car is equal to the change in kinetic energy, which is four times the original kinetic energy.
So, when the speed of a car is doubled, the work to stop the car changes by a factor of 4.