Question

Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 728-kg electric car be able to supply to do the following?

(a) accelerate from rest to 25.0 m/s in 1.00 min
A

(b) climb a 200-m high hill in 2.00 min at a constant 25.0 m/s speed while exerting 443 N of force to overcome air resistance and friction
A

Answer 1

3)gain in energy in 1 min ( 60 sec ) = mv^2 /2

Also, Power required = E/t

= 0.5 mv^2 / t

= 0.5*728*25^2/60

= 3791.67 W

Since efficiency for the conversion of electrical power by the motor = 95.0%

Therefore , P = 0.95 VI

3791.67 = 0.95*12*I

I = 332.6 A

so the current is 332.6 A

b)

height,h = 200 m

L = 120 sec x 25 = 3000 m

Work done,W = 3000 x 443 + 728 x 9.81 x 200 = 2757336 J

P = rate of doing work = W/t = 22977.8

12 * I * 0.95 =22977.8

I = 2015.596 A