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# 1) A roller coaster (398 kg) moves from A (3.82 m above the ground) to B (29.4 m above the ground). Two non-conservative forces are present: friction does -1.82 x 104 J of work on the car, and a chain mechanism does +5.49 x 10^4 J of work to help the car up a long climb. What is the change in the car's kinetic energy KEf - KEo from A to B?

Would I multiply both the distance and work done for both A and B and then subtract them?

Physics(Please respond) - bobpursley, Saturday, June 2, 2012 at 9:39pm
Initial PE-friction+workadded=final PE+final KE

let point A be at Height=0, point B is then (+29.4-3.82)

0-1.82E4+5.49E4=mg(29.4-3.82)+final KE

solve for final KE

For the values of mg, m would be 398kg and g is 9.8 correct?

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1. m would be 398kg and g is 9.8 correct?

Yes, correct

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2. I got 897.45 = 99772.23 + KEf

I subtracted 99772.23 from both sides and got -98874.78 for KEf. Did I do this correctly?

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3. Disregard question. I got the answer.
-63072.23

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