How Steep is the Hill if a Car Uses More Power Climbing It?

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SUMMARY

The discussion focuses on calculating the inclination of a hill based on the power difference experienced by a car traveling up and down. The car, with a mass of 1900 kg, travels at a steady speed of 27 m/s, requiring an additional 49 horsepower (35,000 W) to ascend. The user correctly applies energy principles, determining the height of the hill to be 0.92 m and subsequently calculating the angle of inclination to be approximately 1.95 degrees using trigonometric functions.

PREREQUISITES
  • Understanding of potential energy calculations
  • Basic knowledge of trigonometry, specifically sine functions
  • Familiarity with horsepower and its conversion to watts
  • Concept of energy conservation in physics
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  • Study the principles of energy conservation in mechanical systems
  • Learn about the relationship between power, work, and energy
  • Explore advanced trigonometric applications in physics problems
  • Investigate the effects of friction on energy calculations in motion
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the application of energy principles and trigonometry in real-world scenarios involving motion and inclines.

curly_ebhc
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Homework Statement


There is a car of mass 1900kg that travels at a steady speed of 27 m/s up and down a hill. The force of friction is the same in both directions and it takes 49 more horsepower to go up the hill than down the hill. What is the inclination of the hill.

Given: mass of car= 1900kg , velocity = 27 m/s, difference in power =49 hp =35000 W.

I set it up as an energy problem. First I got rid of time and set that for every second the car traveled 27 m and the difference in the (change in) energy was 35000 J.



Homework Equations





The Attempt at a Solution


Step 1: I set ∆PE(up) -∆PE(down) = 35000J
Then 2∆PE(up)=35000J

Step 2: I used potential energy

h= 35000J/[2(1900kg)(10 m/s2) = .92 m

Step 3: A set up a right triangle: hypotenuse=27m and height=.92m

I used theta=arcsine (.92m/27m) = 1.95 Degrees.

I don’t know the answer but I think that I did everything correctly. Can anyone find a mistake?
 
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Looks right to me!

You might want to give the final answer to 2 sig figs though.

Good job.
 

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