Friction and air resistance of a car travelling down a hill

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Homework Help Overview

The discussion revolves around a physics problem involving a car traveling down a hill, focusing on the concepts of energy loss due to friction and air resistance. The original poster is seeking assistance in calculating the energy lost, given the car's mass, height of the hill, and final speed.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the energy states of the car at the top and bottom of the hill, questioning the initial and final energy calculations. There is an exploration of kinetic energy formulas and the implications of the car's rest state at the top.

Discussion Status

Some participants have provided guidance on the relevant energy concepts and equations, while others are exploring the calculations and questioning their assumptions. There is an ongoing dialogue about the energy transformations involved in the scenario.

Contextual Notes

Participants are encouraged to follow forum rules, including posting relevant equations and showing their thought processes. There is an indication of uncertainty regarding the calculations and the application of energy principles.

lakefall
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A car of mass 1500 kg goes down a hill with a height of 50m. when the car reaches the bottom of the hill the speed is 15 m/s. how much energy was lost due to friction and air resistance?

hey, I am having trouble with my home work T.T I have no clue how to do it if someone can help, much would be much appreciated.

the answer is 566 KJ but i have no clue how to get it.

thanks in advance.
 
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At the top of the incline, how much energy does it have? (remember it is at rest here)

At the bottom how much energy does it have? (remember, relative to the ground, the vertical distance is zero)
 


Hi lakefall and welcome to PF. Please follow the rules of our forum. Use the template for Homework Help, post the relevant equations and show what your thinking is about your question.
 


rock.freak667 said:
At the top of the incline, how much energy does it have? (remember it is at rest here)

At the bottom how much energy does it have? (remember, relative to the ground, the vertical distance is zero)
I'm not sure if I am doing it right, but I used KE = 1/2mv^2 and KE(final) - KE(initial), I think KE(initial) is 0 because you said it is at rest so 0 m/s = V so its 0. But for the KE(final) i got 168750 which is way off haha.

I think I'm doing it wrong. T.T

kuruman said:
Hi lakefall and welcome to PF. Please follow the rules of our forum. Use the template for Homework Help, post the relevant equations and show what your thinking is about your question.
Hi thanks ^_^ okay ill remember to use that template and cute picture.
 


Ok, so you found the kinetic energy at the bottom. Where did that kinetic energy come from?
 

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