Question

A truck engine transmits 28.0 kW (37.5 hp) to the driving wheels when the truck is traveling at a constant velocity of magnitude 60.0 km/h( 37.3mi/h) on a level road. Assume that 65% of the resisting force is due to rolling friction and the remainder is due to air resistance. If the force of rolling friction is independent of speed, and the force of air resistance is proportional to the square of the speed, what power will drive the truck at 30.0 km/h? Give your answer in kilowatts .

Answer 1

Answer:

10319.42 kW

Explanation:

Given:

Total power = 28.0 kW = 28,000 W

Velocity = 60.0 km/hr  = $60\times\frac{\textup{5}}{\textup{18}}$ =  16.67 m/s

65% of the resisting force is due to rolling friction

Force due to air friction, Fₐ ∝ v²

where, v is velocity

or

Fₐ = kv²

here, k is the proportionality constant

Now,

Power = Force × speed

or

28,000 = Resistive force × 16.67

or

Resistive force = 1680 N

thus,

Resistive force due to friction = 0.65 × 1680 = 1092 N

Therefore,

35% of resistive force is due to air friction as 65% is due to friction

thus,

Fₐ = 0.35 ×  1680 N

or

Fₐ = 587.99

also,

Fₐ = kv²

or

587.99 = k × 16.67²

or

k = 2.116

Therefore,

at speed 30 km/hr i.e $30\times\frac{\textup{5}}{\textup{18}}$ = 8.33 m/s

Resistive Force = 1092 N + kv²

= 1092 N + 2.116 × 8.33²

= 1238.827 N

also,

Power = Force × speed

= 1238.827 × 8.33

= 10319.42 kW