# A 1500 kg car has a front profile that is 1.6 m wide by 1.4 m high and a drag coefficient of 0.50. The coefficient of rolling friction is 0.02. What power must the engine provide to drive at a steady 30 m/s if 25% of the power is lost before reaching the drive wheels?

Numerical Question on friction and air drag:A 1500 kg car has a front profile that is 1.6 m wide by 1.4 m high and a drag coefficient of 0.50. The coefficient of rolling friction is 0.02.

What power must the engine provide to drive at a steady 30 m/s if 25% of the power is “lost” before reaching the drive wheels?

**Solution **

The net force on a car moving at a steady speed is zero. The motion is opposed both by rolling friction and by air resistance.

The forward force on the car F_{car} exactly balances the two opposing forces:**F _{car} = f_{r} + F_{drag} **……………………… (1)

Here,

**F**is the drag due to air.

_{drag}

F_{car}= µ_{r}mg + (1/2)CρAv^{2}………………. (2)µ

_{r}= coefficient of rolling friction = 0.02

mass m= 1500 kg

g = 9.8 m/s^{2}

Here A = 11.6 m * 11.4 m is the front cross-section area of the car,

And we used 1.3 kg/m^{3}as the density (ρ) of air.

Drag coefficient = C= 0.50

v = 30 m/sPutting all these values in equation (2) we get

F_{car}= 294 N + 655 N = 949 N

The power required to push the car forward at 30 m/s is

P_{car}= F_{car}x v = 949 x 30 W = 28500 W

This is the power needed at the drive wheels to push the car against the dissipative forces of friction and air resistance. The power output of the engine is larger because some energy is used to run the water pump, the power steering, and other accessories.

In addition, energy is lost to friction in the drive train.

If 25% of the power is lost (a typical value), then P_{car} = 0.75 P_{engine}

So.the engine’s power output P

_{engine}= P_{car}/0.75 = (100/75) x P_{car}= (100/75) x 28500 W = 38,000 W

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