The answer is not 'A'. Using the numbers given in the question, it's 'B'.
It WOULD be 'A' if the number in B were 71 or less. ________________________________
<span>This is not as simple as it looks.
What quantity are we going to compare between the two cases ? Yes, I know ... the "amount of work". But how to find that from the numbers given in the question ? Is it the same as the change in speed ? Well ? Is it ? NO. IT's NOT.
In order to reduce the car's speed, the brakes have to absorb the KINETIC ENERGY, and THAT changes in proportion to the SQUARE of the speed. ( KE = 1/2 m V² )
Case 'A' : The car initially has (1/2 m) (100²) = (1/2m) x 10,000 units of KE.
It slows down to (1/2 m) x (70²) = (1/2m) x 4,900 units of KE.
The brakes have absorbed (10,000 - 4,900) = 5,100 units of KE.
Case 'B' : The car initially has (1/2 m) (79²) = (1/2m) x 6,241 units of KE.
It slows down to a stop . . . NO kinetic energy.
The brakes have absorbed all 6,241 units of KE.
Just as we suspected when we first read the problem, the brakes do more work in Case-B, bringing the car to a stop from 79, than they do when slowing the car from 100 to 70 .
But when we first read the problem and formed that snap impression, we did it for the wrong reason. Here, I'll demonstrate:
Change Case-B. Make it "from 71 km/h to a stop".
Here's the new change in kinetic energy for Case-B:
The car initially has (1/2 m) (71²) = (1/2m) x 5,041 units of KE. It slows down to a stop . . . NO kinetic energy. The brakes have absorbed all 5,041 units of KE.
-- To slow from 100 to 70, the brakes absorbed 5,100 units of KE.
-- Then, to slow the whole rest of the way from 71 to a stop, the brakes absorbed only 5,041 units of KE.
-- The brakes did more work to slow the car the first 30 km/hr than to slow it to a complete stop from 71 km/hr or less.
That's why you can't just say that the bigger change in speed requires the greater amount of work. ______________________________________
It works exactly the same in the opposite direction, too.
It takes less energy from the engine to accelerate the car from rest to 70 km/hr than it takes to accelerate it the next 30, to 100 km/hr !</span>
(The exact break-even speed for this problem is 50√2 km/h, or 70.711... km/hr rounded. )
