A 0.836 kg toy car is powered by two C cells (3.00 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 7.15 s, the car, which is initially at rest, reaches a speed of 1.81 m/s. What is the average current supplied to the car's motor?

  I = 0.160 A

Explanation:

For this exercise we must calculate the power supplied to the motor

power is the work done between the time interval

      P = W / t

      W = F x

      P =

the force is

       F = ma

let's use kinematics

       v = v₀ + a t

as part of rest v₀ = 0

       a = v / t

       a = 1.81 / 7.15

       a = 0.253 m / s

therefore the applied force is

       F = m a

       F = 0.836 0.253

       F = 0.212 N

to calculate the work, let's find the distance traveled

       x = v₀ t + ½ a t²

       x = 0+ ½ a t²

       x = ½ 0.253 7.15²

       x = 6.457 m

therefore The power developed in this movement by the motor

       P = 0.212  6.457/7.15

       P = 0.19145 W

Let's use a direct proportion rule. If the for a supplied power of Ps = 1W to use 0.471W, what power is united when the effective is 0.19145 W

       P_s = 0.19145 W (0.471W/1W)

       P_s = 0.4786 W

       P_s = 0.479 W

how electrical power is

       P = I V

we assume that the batteries are not discharged during the experiment

        I = P / V

electrical power equals supplied power

        P = Ps

calulate

        I = 0.479 / 3.00

        I = 0.1597 A

        I = 0.160 A

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explanation: