Assuming 95.0% efficiency for the conversion of electrical power by the motor, w

Question

Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 823-kg electric car be able to supply to do the following? (a) accelerate from rest to 25.0 m/s in 1.00 min 14.99 95% of the power supplied by the motor goes into increasing the kinetic energy of the car. A (b) climb a 200-m high hill in 2.00 min at a constant 25.0 m/s speed while exerting 513 N of force to overcome air resistance and friction 14.99 In this case, 95% of the power supplied by the motor goes into increasing the potential energy and into the work done against air friction. How is work done related to the applied force and distance traveled? A

Explanation / Answer

(a)

given that,

efficiency of motor e = 95 %

voltage of battery = 12 V

mass of car = 823 kg

speed v = 25 m/s

time t = 1 min = 60 s

We know that,

Power P = E / t

V*l*e = (1/2)mv^2 / t

12 * l *0.95 = (1/2)*823*25^2 / 60

l = 376 A

Current, the battery of electric car able to supply = 376 A

(b)

As given , height h = 200 m

t = 2 min = 120 s

v = 25 m/s

F = 513 N

We know that,

Power P = E / t and P = F*v

so Here,

V*l*e = (F*v) + (mgh / t)

12 * l *0.95 = (513*25) + (823*9.8*200 / 120)

l = 2305.35 A

Current, the battery of electric car able to supply = 2305.35 A

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.