A 500 kg car slows down due to friction from 30 m/s to 20 m/s. What was the work done against friction. ?
- Demiurge42Lv 71 month ago
I don't know what you mean by "work done against friction".
The net work done on the car will equal the car's change in kinetic energy.
Kinetic energy: KE = 0.5 mv^2
The change in kinetic energy of the car is:0.5*500kg*(20m/s)^2 - 0.5*500kg*(30m/s)^2 = -125,000J
If friction is the only force slowing the car then the work done by friction is -125,000J
- billrussell42Lv 71 month ago
v² = v₀² + 2ad
30² = 20² + 2ad
900 – 400 = 2ad
ad = 250
a = 250/d
F = ma = 500•250/d = 125000/d
work = Fd = d(125000/d) = 125000 J