Question

A car of mass 2 000 kg and engine power 320 HP = 240 kW is climbing an incline with sin(θ)=0.04 at speed a speed of 180 km/hr (50 m / s). Ignore all losses except air drag which is proportional to the square of the speed of the car (FD∝v2). Take g=10m/s2

PART A How much power the engine is exerting doing against gravity?
80 kW
200 kJ
120 kW
160 kJ
40 kW

PART B Assume that the rest of engine power is exerted against air resistance. How much is this power?
a) 400 kW
b) 200 kW
c) 1 MJ
d) 800 kJ
e) 600 kW

PART C What is the air resistance force at this speed?
a) 20 kW
b) 4 kN
c) 16 kJ
d) 12 kN
e) 8 kN
PART D If the same car is moving on flat ground at 90 km / hr = 25 m/s how much power would it exert?
a) 25 W
b) 50 kW
c) 25 kW
d) 60 W
e) 100 W

Answer 1

The drag force acting on the car increases as the speed of the car

increases.

The correct values are;

• PART A; The power of engine against gravity is 40 kW

• PART B; Engine power exerted against the air is b) 200 kW

• PART C; The air resistance force is b) 4 kN

• PART D The power the car exerts is c) 25 kW

Reasons:

The given parameters are;

Mass of the car = 2,000 kg

Engine power, P = 320 Hp = 240 kW

Inclination of the direction of motion, sin(θ) = 0.04

Speed of the car, v = 180 km/hr (50 m/s)

Losses = Air drag $F_D$ ∝ v²

PART A Power engine exerts against gravity;

The gravitational force acting along the plane$F_{grp}$ = m·g·sin(θ)

The power of engine against gravity, $P_{grp}$ = $F_{grp}$ × v

Therefore;

$P_{grp}$ = 2000 kg × 10 m/s² × 0.04 × 50 m/s = 40,000 W = 40 kW

The power of engine against gravity, $P_{grp}$ = 40 kW

PART B Engine power exerted against the air;

Let $P_{pa}$ represent the engine power exerted against the air, and we have;

P = $P_{pa}$ + $P_{grp}$

Therefore;

$P_{pa}$ = P - $P_{grp}$

Which gives;

$P_{pa}$ = 240 kW - 40 kW = 200 kW

Engine power exerted against the air, $P_{pa}$ is b) 200 kW

PART C Air resistance force at the given speed

Let $F_a$ represent the air resistance force, we have;

$Force =\mathbf{ \dfrac{Power}{Speed}}$

Therefore;

$F_a = \dfrac{P_{pa}}{v}$

Which gives;

$F_a = \dfrac{200 \, kW}{50 \, m/s} = 4 \, kN$

The air resistance force, $F_a$ is b) 4 kN

PART D Power exerted by car moving on flat ground

$F_D$ ∝ v²

$F_D$ = k·v²

4 kN = k × (50 m/s)²

$k = \dfrac{4 \, kN}{(50 \, m/s)^2} = 1.6 \, kg/m$

The drag force when the car is moving at v = 25 m/s, $F_D$ is therefore;

$F_D$ = 1.6 kg/m × (25 m/s)² = 1,000 N

Power, P = Force × Speed

Therefore,  P = $F_D$ × v

P = 1,000 N × 25 m/s = 25 kW

The power the car exerts, P is c) 25 kW

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