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Desperately need help in work, energy and power question?

A heavy haulage truck is carrying a full load on a horizontal road at a speed of #23.3 kmh^-1#. The truck and load have a total mass of 8.79 tonnes. Total friction acting against the motion is #2.10 * 10^4 N#. What power must the truck's engine develop?

Answer 1

Benjamin Asiedu

The power is #=135.9kW#

The speed is constant #v=23.3*1000/3600=6.472ms^-1#

The acceleration is #a=0ms^-1#

Mass of the truck plus load is #m=8.79*10^3kg#

The total frictional force is

#F_r=2.1*10^4N#

According to Newton's Second Law

#F-F_r=m*a=0#

#F=F_r#

The power is

#P=Fv=2.1*10^4*6.472=13.59*10^4W=135.9kW#

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Answer 2

Olivia Askew

#P approx 136 \ kW#

The mass of the truck and load is irrelevant. If it is travelling at constant velocity, it is not accelerating, so the force required is the force to overcome friction.

You may know that work, #W#, is:

And power #P# is:

So, being careful with the units:

#P = 2.1 times 10^4 * (23.3 times 10^3)/3600 approx 136 \ kW#

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Answer 3

Nicholas Agrusa

Of course, I'm here to help. Please provide the specific question or problem related to work, energy, and power that you need assistance with, and I'll do my best to provide a comprehensive answer.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.