A roller coaster (420 kg) moves from A (5.00 m above the ground) to B (18.0 m above the ground). Two nonconservative forces are present: friction does -2.00 104 J of work on the car, and a chain mechanism does +3.00 104 J of work to help the car up a long climb. What is the change in the car's kinetic energy, KE = KEf - KE0, from A to B?
2007-07-30 13:40:24 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
This is a conservation of energy problem. The total initial energy of the coaster (gravitational potential plus kinetic) plus all work done on the coaster must equal the total final energy of the coaster. If you use the ground as height equals zero for the formula GPE = mgh, gravitational potential energy equals mass times gravity time height, you can calculate initial and final GPE. You don't need to calculate KEf or KE0. You can call the work done by friction F and the work done by the chain W. Then you have GPE0 + KE0 + F + W = GPEf + KEf. Remember that F and W are given (and F has a negative value) and I've shown you how to calculate GPE, so you'll have an equation that you can rearrange algebraically to get a value for KEf - KE0.
2007-07-30 13:47:24 · answer #1 · answered by DavidK93 7 · 0⤊ 0⤋
your final KE is zero .. because at the end car stops.... now the total energy used is Friction+mechanicalE.. which is
(-2.00104J + 3.00104J).. so now total change in KE = KEf -KEo = 0 - (friktion plus mechanical)
2007-07-30 20:50:27 · answer #2 · answered by xreighted s 2 · 0⤊ 0⤋
is it 13 M times 420kg??
2007-07-30 20:47:19 · answer #3 · answered by ebiz1@sbcglobal.net 2 · 0⤊ 1⤋