Max Power of a Tesla SUV: Solving F(net)=F(app)-F(resistance)

[physicslady123]

Homework Statement

A New Tesla SUV with a mass of 1.5*10^3 kg starts from rest and accelerates to a speed of 100 km/h in 6s. Assume that the force of resistance remains constant at 600N during that time. Qhat is the average power delivered by the car's engine?

Relevant Equations

P=ΔE/Δt
W=Fd

F(net)=F(app)-F(resistance)
ma=F(app)-F(resistance)
(1.5*10^3)(16.67)=F(app)-600
25005=F(app)-600

P=ΔE/Δt
P=25005/6
P=4167.5 W

I got the incorrect answer (because I didn't take into account the resistance?). I am also confused about which Force value to use for the "E" value. Do I use the Force(applied) or the Force(net)?

 

[haruspex]

I got the incorrect answer (because I didn't take into account the resistance?).

Of course. You calculated the engine power that would have been needed if there were no resistance, so the actual engine power must have been more.

I am also confused about which Force value to use for the "E" value. Do I use the Force(applied) or the Force(net)?

You want the total power from the engine, not just the power that went into producing the change in KE.

Btw, the speed is given in km/h. Always keep track of units in your work.

 

[haruspex]

Incidentally, I think there is a flaw in the question. It does not specify that acceleration is constant. You have assumed it is, and I believe you need to make some such assumption or there is insufficient information.
I'll check this, but I don’t have time right now.

Edit: have now confirmed my suspicion. E.g.

1 Suppose all the acceleration happens in the first fraction of a second, so it then travels at speed v for the rest of the time t.
Distance covered=vt
Work against friction =vtF
KE gain = ½mv²
Work done by engine= vtF+½mv²
Average power=$vF+\frac{mv^2}{2t}$

2. Suppose all the acceleration happens in the last fraction of a second. It spent most of the time t stationary.
Distance covered=0
Work against friction =0
KE gain = ½mv²
Work done by engine= ½mv²
Average power=$\frac{mv^2}{2t}$