A truck engine transmits 29.5 kW (39.5 hp ) to the driving wheels when the truck

Question

A truck engine transmits 29.5 kW (39.5 hp ) to the driving wheels when the truck is traveling at a constant velocity of magnitude 60.0 km/h (37.3 mi/h ) on a level road.

A. What is the resisting force acting on the truck?

ANSWER: 1770N

B. Assume that 64 % 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 25.0 km/h ? Give your answer in kilowatts .

ANSWER: 8.63 kW

C. Assume that 64 % 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 25.0 km/h ? Give your answer in horsepower.

D. Assume that 64 % 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 112.0 km/h ? Give your answer in kilowatts.

E. Assume that 64 % 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 112.0 km/h ? Give your answer in horsepower.

Explanation / Answer

Hi,

Question A: We need to convert the units to just one system, so in the international system the power must be in watts (W) and the speed must be in m/s

power=29,5kW*1000W/1kW=29500W

speed=(60km/h)*(1h/3600s)*(1000m/1km)=16.6666m/s

As the speed is constant, Power = Fr x speed where Fr: resisting force

so the total resisting force Fr= power/speed=29500W/16.666m/s=1770N

Question B: The 64% of 1770 is: 1770*64/100=1132.8N=Frf:force due to rolling friction

Force due to air resistance:Fa=1770N-1132.8N=637.2N

Air resistance force is proportional to the square of speed, Fa = k v^2.
Fao = 637.2N = k x (16.7)^2
k = 2.28
now if the speed is reduced to 25km/h, speed=(25km/h)*(1h/3600s)*(1000m/1km)=6.944m/s

then Fa = 2.28 x (6.944)^2 = 110.17N

The total resistive force is now Frf + Fa = 1132.8N+110.17N=1243N

So the total drive power is 1243Nx 6.944m/s = 8.63kW

Question C:

power=8.63kW*(1.34hp/1kW)=11.56hp

Question D: very similar to question B the only difference is the speed=(112km/h)*(1h/3600s)*(1000m/1km)=31.11m/s

from the question B, we have: Frf=1132.8N, Fao=637.2N, k=2.28

then Fa = 2.28 x (31.11)^2 = 2206.7N

The total resistive force is now Frf + Fa = 1132.8N+2206.7N=3339.46N

So the total drive power is 3339.46Nx 31.11m/s = 103.9kW

Question E:

power=103.9kW*(1.34hp/1kW)=139.218hp

I hope this will help you =)

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