Power Calculation for a Car Driving on a Hilly Road with a Given Slope

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Discussion Overview

The discussion revolves around calculating the power delivered by a car's engine while driving up a hilly road with a specified slope. Participants explore the application of energy equations in the context of a physics homework problem, focusing on the relationship between kinetic and potential energy and the concept of power.

Discussion Character

  • Homework-related, Mathematical reasoning, Technical explanation

Main Points Raised

  • One participant presents a problem involving a car of mass 850 kg driving at 70 km/hr up a 30° slope and attempts to use the macroscopic energy equation to calculate the power.
  • The participant expresses confusion about the presence of an unknown time variable (t) in their power calculation.
  • Another participant suggests that the original poster should consider variations in energy rather than a static equation for power.
  • A further reply emphasizes the importance of understanding the relationship between energy changes and time, proposing the use of instantaneous power or changes in energy over time.

Areas of Agreement / Disagreement

Participants do not reach a consensus on how to approach the problem, with differing views on the correct application of energy equations and the treatment of time in the power calculation.

Contextual Notes

The discussion highlights potential limitations in the original approach, such as the dependence on the definition of power and the need for clarity regarding the time variable in the context of energy changes.

Who May Find This Useful

This discussion may be useful for students working on similar physics problems involving energy calculations, power, and the dynamics of vehicles on inclined surfaces.

LauraMorrison
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Homework Statement



A car of mass 850 kg is driven at a steady speed of 70 km/hr up a hilly road
with a slope of 30°. Using the macroscopic energy equation, determine the
power delivered by the engine of the car.

Homework Equations


E= KE + PE
E = (1/2)mC^(2) + mgz
Power = E/t

The Attempt at a Solution



My attempt at a solution was:
E = 1/2mC^(2) + mg(Ctsin(30))
E = 1/2(850)(19.44)^(2) + (850)(9.81)(19.44)tsin(30)

Therefore

Power = (1/2(850)(19.44)^(2))/t + (850)(9.81)(19.44)sin(30)

Why do I still have an unknown t? I am supposed to be able to solve it with the information given but I can't get it!
Please please help!
 
Last edited:
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Your problem is with
LauraMorrison said:
Power = E/t
You should be considering variations...
 
Do you mean the variations in potential and kinetic energy as the car drives up the hill? I am not sure how to calculate that, would you be able to explain?

I am sorry, I know it is such a simple question.
 
If you write ##P = E/t##, think what happens if the car is going at constant speed on a flat road.

The power is used to change the energy of the car, so you have to consider ##P = \Delta E / \Delta t## or, even better, the instantaneous power ##P = dE / dt##.

Hope this helps. Don't hesitate with further questions if it doesn't!
 

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