Two cars have the same mass, but one is moving three times faster than the other. How much more work will be needed to stop the faster car?
KE = (1/2) m v^2
You must do enough work (Force times braking distance) to equal the original Kinetic energy.
W1 = (1/2) m v^2
W2 = (1/2) m (3v)^2 = (1/2) m v^2 * 9
so
nine times as much
I guess NINE TIMES AS MUCH
To calculate the work needed to stop the faster car, we need to consider the kinetic energy of the cars. The kinetic energy is given by the equation:
K.E. = (1/2) * m * v^2
where m is the mass of the car and v is its velocity.
Let's assume the mass of both cars is m.
For the slower car:
K.E.1 = (1/2) * m * v1^2
For the faster car:
K.E.2 = (1/2) * m * v2^2
Since the mass of both cars is the same, m is cancelled out in the equations.
Given that the faster car is moving three times faster than the slower car, we can write v2 = 3 * v1.
Substituting this value in K.E.2, we get:
K.E.2 = (1/2) * m * (3 * v1)^2 = (1/2) * m * 9 * v1^2 = 4.5 * m * v1^2
To find the difference in work needed to stop the faster car compared to the slower car, we subtract K.E.1 from K.E.2:
Work = K.E.2 - K.E.1 = 4.5 * m * v1^2 - (1/2) * m * v1^2 = 3.5 * m * v1^2
Therefore, to stop the faster car, 3.5 times more work will be needed compared to stopping the slower car.
To calculate the work needed to stop a car, we can use the equation:
Work = (1/2) * mass * velocity^2
Given that the two cars have the same mass, we can assume it as 'm'. Let's say the slower car has a velocity 'v' and the faster car has a velocity '3v'.
The work required to stop the slow car can be calculated as:
Work_slow = (1/2) * m * v^2
The work required to stop the fast car can be calculated as:
Work_fast = (1/2) * m * (3v)^2
Simplifying:
Work_fast = (1/2) * m * 9v^2
Now, to find out how much more work is needed to stop the faster car, subtract the work required for the slow car from the work required for the fast car:
Difference in work = Work_fast - Work_slow
Difference in work = (1/2) * m * 9v^2 - (1/2) * m * v^2
Difference in work = (1/2) * m * (9v^2 - v^2)
Difference in work = (1/2) * m * (8v^2)
Therefore, the amount of additional work needed to stop the faster car is (1/2) * m * (8v^2).