Friction and air resistance of a car travelling down a hill

Click For Summary
SUMMARY

A car with a mass of 1500 kg descends a hill with a height of 50 meters, reaching a speed of 15 m/s at the bottom. The total energy lost due to friction and air resistance is calculated to be 566 kJ. The initial potential energy at the top of the hill is converted into kinetic energy at the bottom, but some energy is dissipated due to friction and air resistance. The kinetic energy at the bottom is calculated using the formula KE = 1/2 mv², resulting in a final kinetic energy of 168750 J.

PREREQUISITES
  • Understanding of gravitational potential energy (PE = mgh)
  • Knowledge of kinetic energy calculations (KE = 1/2 mv²)
  • Familiarity with energy conservation principles
  • Basic algebra for solving equations
NEXT STEPS
  • Study the principles of energy conservation in physics
  • Learn about the effects of friction and air resistance on moving objects
  • Explore advanced kinetic energy calculations in different scenarios
  • Investigate real-world applications of potential and kinetic energy in automotive engineering
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of energy loss due to friction and air resistance in practical scenarios.

lakefall
Messages
2
Reaction score
0
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.
 
Physics news on Phys.org


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?
 

Similar threads

Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
2
Views
5K
  • · Replies 45 ·
2
Replies
45
Views
10K
  • · Replies 3 ·
Replies
3
Views
2K