Work & energy

[circulus vitios]

What would the final speed of a car be if fuel efficiency were doubled. Neglect air drag.

A. v
B. sqrt(2)v
C. 2v
D. 4v

Could someone write out the algebraic proportionality? It makes intuitive sense but I'm having a brain-dead moment and I can't figure out how to go from W=ΔKE to the answer.

---

Members don't see this ad.

 

[milski]

Wdone = efficiency * Wtotal
From here,
ΔKEbe=2ΔKEwe (be - better efficiency, we - worse efficiency)
KE=mV^2/2 or V=sqrt(2KE/m)
Vwe = sqrt(2KE/m)=sqrt(2KEwe/m)
Vbe = sqrt(2KE/m)=sqrt(2KEbe/m)=sqrt(4KEwe/m)=sqrt(2)*sqrt(2KEwe/m)=sqrt(2)*Vwe

How's that? Formatting is rather crappy but it should be clear enough.
The only assumption that's not in what you quoted (but should be in the question) is that the car starts from 0 velocity and the KEfinal=ΔKE-0=ΔKE

 

[circulus vitios]

Wdone = efficiency * Wtotal
From here,
ΔKEbe=2ΔKEwe (be - better efficiency, we - worse efficiency)
KE=mV^2/2 or V=sqrt(2KE/m)
Vwe = sqrt(2KE/m)=sqrt(2KEwe/m)
Vbe = sqrt(2KE/m)=sqrt(2KEbe/m)=sqrt(4KEwe/m)=sqrt(2)*sqrt(2KEwe/m)=sqrt(2)*Vwe

How's that? Formatting is rather crappy but it should be clear enough.
The only assumption that's not in what you quoted (but should be in the question) is that the car starts from 0 velocity and the KEfinal=ΔKE-0=ΔKE

Yup, you're right. Thanks!