Question

Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 678-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 353 N of force to overcome air resistance and friction A

Answer 1

Then energy change the ņ xVxIxt = { m (Vx" v%) here, m = ergy potential energy and frictions dissipated by engine must be equal to in kinetic ener energy of car. nie. efficiency, na 954. = 0.95 voltage v=12V Current I .? t 1min = 60 sec 678 kg VF = 25 m/s Vi = o mis (Asit start from rest) + m (vg-vo) nut 678* 252 0.95412 x 60 619.5 ~ 620 A (b). Now when with constant velocity and over come friction then, total electrical energy by the kinetic ener energy nvit Imvet night tFh :: I = - car move dissiputa mw²t might fh :..I = nut

x 678X2534 678x9.8x 200 + 35 3x 200 i. Ia m=678kg g=9.8 2012 h=200m 0.95X 12 X 2 X 60 ta_2x60 sec F=353N - 1177.89 17 1178 We know, Resistance, R= p. l A vesistivity (P1= 1.72x10 82.m. length (l) = 208 x 1000 m ror=rx 00055 Sechim (A) x( ? 9: 2668103 xios)? 2 8 2800 IR = 1.72x10 Rx (9.26C xid?)? -0.714 e iR .. No power loss = = (167) *x0.714 = 19912.746 W.