Question

A 0.954-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 36.1%, meaning that 36.1% of the electric energy applied to the motor is converted into translational kinetic energy. After 7.88 seconds, the car, which is initially at rest, reaches a speed of 1.27 m/s. What is the average current supplied to the car\'s motor?

Answer 1

0.06 A

Step-by-step explanation:

Given:

Mass of the toy car, m = 0.954 kg

Total potential provided, V = 4.50 volts

Efficiency of the toy, η = 36.1 % = 0.361

Time, t = 7.88 seconds

Initial speed of the toy car, v = 1.27 m/s

Now, the energy being released by the car while running is the kinetic energy

thus,

E = (1/2)mv²

on substituting the values, we get

E = (1/2) × 0.954 × 1.27²

or

E = 0.769 J

also,

Energy = Power × time

or

Power = Energy/time

on substituting the values, we get

Power = 0.769/7.88 = 0.0976 W

now,

the efficiency is given as;

η = (output power) / (Input power)

Input power = Voltage × current (I)

or

Input power = 4.50 × I

thus,

0.361 = 0.0976/(4.5 × I)

or

I = 0.060 A

Answer 2

0.06 A

Step-by-step explanation:

We have given mass =0.954 kg

velocity =1.27 m/sec

efficiency =0.361 the output kinetic energy of the motor
`KE=(1)/(2)mv^2=(1)/(2)* 0.954* 1.27^2=0.769\ J`

Efficiency =0.361

so input to the motor = output/efficiency

so input to the motor =
`(0.769)/(0.361)=2.1301`

we know that
`P=(E)/(T)` where E is energy and T is time so
`P=(2.1301)/(7.88)=0.2703W`

We know that power P=VI we have given V=4.5 VOLT

So current
`I=(P)/(V)=(0.2703)/(4.5)=0.06 A`