Drag Force and Power on an inclined plane

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Homework Help Overview

The problem involves calculating the power required by a car to overcome drag forces while driving at a constant speed on both level ground and an inclined plane. The subject area includes dynamics and forces, specifically focusing on drag force and power calculations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the application of the power formula P=Fv, with initial attempts to calculate power for both scenarios. Some participants raise questions about the effects of slope on drag force and suggest using trigonometric relationships to find sin(theta) from the given slope.

Discussion Status

There are multiple interpretations being explored regarding the calculation of power on an incline. Some participants have provided calculations and expressed uncertainty about the results, while others are discussing the necessary adjustments to account for the incline's effect on drag force.

Contextual Notes

The original poster expresses confusion about the problem, and there are references to using trigonometric functions to derive necessary values, indicating that assumptions about the slope and drag forces are under discussion.

kraaaaamos
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Homework Statement



A 710 kg car drives at a constant speed of 23 m/s. It is subject to a drag force of 500N. What power is required from the car’s engine to drive the car
a. on a level ground?
b. up a hill with a slope of 2.0 0?


Homework Equations



P=Fv


The Attempt at a Solution



a) P = (500)(23)?

b) P = (500cos2)(23)?

I HAVE NO IDEA.
 
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In the second part slope is given. Slope = tan(theta). From that find sin(theta). Now the total drag force during up hill drive = 500 + Mgsin(theta)
 
so for b
its:


P = (500+ mgsin(theta))(v)
= [500 + (710)(9.8)(sin2) ](v)
= 17 085 W?
 
P = (500+ mgsin(theta))(v)
= [500 + (710)(9.8)(sin2) ](v)
= 17 085 W?

Tan(theta) = 2. Using trigonometrical tables find theta and then find theta or if Tan(theta) = 2. the sin(theta) = (sqrt.5)/2 Use this and find the value.
 

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