Answer
See answers.
Work Step by Step
a. The coils in this car are directly connected to the wheels with no transmission. Therefore the motor’s required torque, using equation 20-10, equals the torque caused by the frictional force.
$$\tau=NIAB=Fr$$
Solve for the current.
$$I=\frac{Fr}{NAB}=\frac{(250N)(0.29m)}{290(0.12m)(0.15m)(0.65T)}\approx 21A$$
b. By energy conservation, the power lost due to friction equals the net power provided by the coils. Calculate the latter as the current multiplied by the back emf.
$$P=Fv=I\epsilon_{back}$$
$$ \epsilon_{back}=\frac{Fv}{I}=\frac{(250N)(35000m/3600s)}{21.368A}\approx 110V$$
c. The power dissipated in the coils is the power produced by the coils, minus the net power provided to the wheels. Ten 12-V batteries in series is 120 volts.
$$P_{lost}=P-P_{wheels}=21.368A(120V-113.75V)\approx 130 W$$
d. To find the fraction of input power used to drive the car, divide the net power provided to the wheels by the total power.
$$\frac{P_{wheels}}{P}=\frac{I(113.75V)}{I(120V)}=0.9475\approx 95\%$$