Question

Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 788-kg electric car be able to supply to do the following?

(a) accelerate from rest to 25.0 m/s in 1.00 min Correct: 359.4 A Your answer is correct.

(b) climb a 200-m high hill in 2.00 min at a constant 25.0 m/s speed while exerting 503 N of force to overcome air resistance and friction A

I got (a) I just need help with (b). Please show work so I understand.

Answer 1

(b)
Distance = 25.0 * 120 = 3000 s
F = mgsin(θ)
F = (788*9.81)* (200/3000) = 515.3 N

Power = Total Force*const speed
P = (515.3 + 503 ) * 25
P = 25457.5 W

We know,
P = V * I
As efficency is 95%
P = 0.95 * 12.0 * I
I = 25457.5/(12*0.95)
I = 2233 A