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 70 km/h to a stop".
Here's the new change in kinetic energy for Case-B:
The car initially has (1/2 m) (70²)
= (1/2m) x 4,900 units of KE.
It slows down to a stop . . . NO kinetic energy.
The brakes have absorbed all 4,900 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 70 to a stop,
the brakes absorbed only 4,900 units of KE.
-- The brakes did more work to slow the car the first 30 km/hr
than to slow it the whole remaining 70 km/hr.
That's why you can't just say that the bigger change in speed
requires the greater amount of work.
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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 !
