Forces and Power: 18.5 HP to Drive 2320 kg Auto at 62.0 km/h

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SUMMARY

The discussion focuses on calculating the forces and power required to drive a 2320 kg automobile at 62.0 km/h, specifically addressing two parts: the total retarding force and the power needed to ascend a 15% grade. The calculations establish that 18.5 HP equates to 13795 W, leading to a retarding force of approximately 0.3498 N on a level road. For the uphill scenario, the required power to overcome gravitational forces and maintain speed is calculated to be 77 HP, correcting initial miscalculations by applying the appropriate formulas for force and power.

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Homework Statement


(A) If 18.5 HP are required to drive a 2320 kg automobile at 62.0 km/h on a level road, what is the total retarding force due to friction, air resistance, and so on?
(B) What power is necessary to drive the car at 62.0 km/h up a 15% grade (a hill rising 15.0m vertically in 100m horizontally).

The Attempt at a Solution


18.5 HP = 13795 W
62.0 km/h = 17 m/s
A) [tex]P = F*v[/tex]
[tex]P / v = F[/tex]
[tex]F = ma[/tex]
[tex]P / v = ma[/tex]
[tex]a = P / (v * m)[/tex]
[tex]13795 / (17 * 2320) = 0.3498N[/tex]
B) [tex]\alpha = arctan(15/100) = 8.53[/tex]
[tex]\sum F = F_{up} - 0.3498N - w sin(8.530)[/tex]
[tex]\sum F = F_{up} - 0.3498N - 3375N[/tex]
[tex]F_{up} = 3375N[/tex]
[tex]P = F * v[/tex]
[tex]P = 3375 * 17 = 57375 W = 77 HP[/tex]

This seems completely wrong to me... I'm not sure if any of it is right even. Any help would be greatly appreciated.
 
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Your part a) isn't right. You had the right formula, but didn't use it. F = P/v.

plug in P plug in v. get F.

Part b), your work is right... you just need to use the correct force from part a).
 
Thanks learningphysics! I had no idea I was so close to the answer, I figured all my work was wrong :(
 

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