Power Required to Climb 10% Grade at 35 m/s with 600N Force

[spj1]

Car travels up 10% grade at constant speed of 35 m/s. The car is 2000 kg and requires a total force of 600N on a regular, flat incline. What is the power required to get up the grade?




P=FxV




%10 percent incline = arctan(.1)... theta = 5.7, so the displacement would be about 10.1 (depending how you treat the grade). Then I plugged into the equation P = Fcos(theta)(displacement)(velocity)... This does not yield the correct answer! It's got to be how I'm dealing with the grade. I'm not sure what to do with it. Any hints?

 

[PhanthomJay]

I'm guessing from the problem that a 'flat incline' is one with no slope, and that it requires 600N force to keep the car moving at 35m/s on a horizontal surface (apparently to overcome friction and air drag, etc.). So when it moves up the incline, it must provide 600N PLUS an additional component to overcome the gravity component acting down the plane. No need to look at displacements, just calculate the required 'F' and multiply it by V. Your angle theta is correct.

 

[Moetasim]

There are a few potential issues with the calculation that could be leading to an incorrect answer. First, it is important to clarify what is meant by a "regular, flat incline". If the incline is truly flat, then the force of 600N would not be necessary to maintain a constant speed of 35 m/s. However, if the incline is meant to be a constant 10% grade, then the force required to maintain a constant speed would be greater than 600N.

Additionally, the equation P = Fcos(theta)(displacement)(velocity) may not be applicable in this situation. This equation is typically used to calculate the power required to overcome a constant force and move an object a certain distance at a constant velocity. However, in this scenario, the car is already moving at a constant velocity and the force is not constant due to the changing grade of the road.

To accurately calculate the power required, it may be necessary to use a different equation, such as P = Fv, where F is the total force required to overcome the grade and maintain a constant velocity of 35 m/s, and v is the velocity of the car.

It is also important to consider the units being used in the calculation. The force is given in Newtons, but the velocity is given in meters per second. These units would need to be converted to a common unit, such as watts, in order to accurately calculate the power required.

In conclusion, to accurately determine the power required to climb a 10% grade at 35 m/s with a force of 600N, more information and clarification is needed on the specifics of the scenario and the equations being used. It may also be helpful to consult with a physics expert for assistance in solving this problem.