A car has a mass of 0.098 kg and an initial velocity of 2.2 m/s. If h1 is 0.50 m and h2 is 0.30 m, calculate the mechanical energy, in joules, of the car at the low point before hill a.
I tried a similar q but it's totally wrong. please help!!
here's the link:http://www.flickr.com/photos/62137592@N03/5700213813/in/photostream/
I tried a similar q but it's totally wrong. please help!!
here's the link:http://www.flickr.com/photos/62137592@N03/5700213813/in/photostream/
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kinetic energy = 1/2 m v^2
potential energy = mgh
add them to get total energy at any point.
If the car is going downhill, add the LOSS of potential energy to the original kinetic to get the new kinetic energy,
= 1/2* 0.098 * 2.2 ^2 + 0.098 * 9.8 * (0.5 - 0.3)
(= 0.43 approx Joules)
potential energy = mgh
add them to get total energy at any point.
If the car is going downhill, add the LOSS of potential energy to the original kinetic to get the new kinetic energy,
= 1/2* 0.098 * 2.2 ^2 + 0.098 * 9.8 * (0.5 - 0.3)
(= 0.43 approx Joules)
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If we assume there are no losses due to friction, then the mechanical energy of the car at the bottom of the hill will be the same as it was at the top of the hill. Because the car is at a lower elevation, it has LESS potential energy, but it will have MORE kinetic energy because it will be accelerating as it rolls downhill. The sum of these is the mechanical energy. So you need to calculate the potential energy and kinetic energy at the top of the hill, add them together, and that is the mechanical energy (at h1).
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IF you don't know it by now, you'll not know it in time for the test.