Question

A 0.482 kg toy car is powered by three D cells (4.50 V total) connected directly to a small DC motor. The car has an effective energy conversion efficiency of 47.1 % , meaning that 47.1 % of the electric energy applied to the motor is converted into translational kinetic energy. After 7.88 s , the car, which is initially at rest, reaches a speed of 2.35 m / s . What is the average current supplied to the car's motor?

Answer 1

Step 1: Using Work-energy theorem to find Work-done required by the batteries to reach final given speed:

W = dKE

W = KEf - KEi

W = (1/2)*m*Vf^2 - (1/2)*m*Vi^2

m = mass of car = 0.482 kg

Vf = final speed of car = 2.35 m/s

Vi = Initial speed of car = 0 m/s

So, W = (1/2)*0.482*2.35^2 - 0 = 1.33 J

Now given that car has effective energy conversion efficiency of 47.1%, So

W_net = Net work-done by car = W/n

n = efficiency = 47.1% = 0.471

W_net = 1.33/0.471 = 2.824 J

Step 2: Now we know that Relation between Power and energy is given by:

P = W_net/t

P = Power Supplied by batteries = V*I

V*I = W_net/t

I = W_net/(V*t)

V = Voltage supplied = 4.50 V

t = time interval = 7.88 sec

So,

I = Current

I = 2.824/(4.50*7.88)

I = 0.0796 Amp = Current supplied

Let me know if you've any query.