(Work, Power, Energy Question)?

Question

A car working at a rate 500W has a maximum speed 20 m/s when travelling on the level against resistance proportional to the square if its speed. At what rate would the car have to work to double its maximum speed?

Answer 1

P = 500W
v = 20m/s
let F be the force of the car and f be the force of resistance
then P = Fv that is F= P/v
(where P = Fv should be in your textbook)
and by supposition
f = kv^2
where k is the constant of proportionality
then, since there is no acceleration, the force of the car is equaled by the resistance force, that is
F = f
so
P/v = kv^2
gives
P = kv^3
so the power is proportional to the velocity cubed
next
let P' be the new power required to travel
at v' = 2v twice the original velocity
then
P' = kv'^3= k(2v)^3 = k(2^3)v^3 = 8kv^3= 8P
so
P' = 8*500 ≈ 4000W

note the v = 20m/s was useless information unless you were interested in finding out what k equals

the end
.

Answer 2

If the frictional forces are a squared function of its speed, it would have to produce 2000 W (2 kW) to travel at twice its speed. That is, it would require four times the energy (4 * 500 W = 2 kW)

Doug