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

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To calculate the power required for a car to climb a 10% grade at 35 m/s, the total force needed is 600N plus an additional force to counteract gravity. The incline creates an additional gravitational component that must be factored into the total force. The correct approach involves using the formula P = F * V, where F is the total force and V is the velocity. The angle of the incline, approximately 5.7 degrees, is relevant for determining the gravitational force component but does not require displacement calculations. Thus, the power required is the product of the total force and the constant speed of the car.
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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?
 
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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.
 
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