A 1310 kg car drives up a hill that is 16.5 m high. During the drive, two nonconservative forces do work on the car: (i) the force of friction, and (ii) the force generated by the car's engine. The work done by friction is -3.21 x 10^5 J; the work done by the engine is +6.33 x 10^5 J. Find the change in the car's kinetic energy from the bottom of the hill to the top of the hill.
Thanks so much for your help!!! God Bless!!
2007-11-01 08:56:51 · 3 answers · asked by PaNaSoNiCeDgE 1 in Science & Mathematics ➔ Physics
What would be the actual nuumerical answer for this problem???.......I cant seem to get it right. Thanks!!
2007-11-01 09:36:16 · update #1
I wish I wuz one. The genius I mean...
I think that...
Ke=0.5mV^2
Ke(bottom of the hill)=0.5x 1310 V^2
Work (against gravity) = m g h=
Work (against gravity)=1310 x 9.81x 16.5=
Work (against gravity)=2.12 x 10^5 J
also
Ke(top of the hill)= Work(done by engine) - Work (against friction) - Work (against gravity)
Ke(top of teh hill)=6.33 x 10^5 - 3.21 x 10^5 - 2.12 10^5 =1.0x 10^5J
DKe = Ke(top of the hill) - Ke(bottom of the hill)
DKe= 1.0x 10^5 -0.5x 1310 V^2
so what was the speed of the car at the bottom of the hill?
2007-11-01 09:05:36 · answer #1 · answered by Edward 7 · 1⤊ 0⤋
Non Conservative work = Change in Mechanical Energy
Wnc = Ef - Ei
Ei = KEi + PEi we can assume PEi =0 at bottom of hill
Ef = KEf +PEf and PEf = mgh, where h =16.5m
The non conservative Work is
Wnc = Wengine + Wfriction
so
Wengine + Wfriction = KEf + mgh - KEi
you want the change in KE, so solve for KEf- KEi
Don't forget the negative sign for Wfriction!
2007-11-01 09:07:24 · answer #2 · answered by Anonymous · 0⤊ 1⤋
-3.21 x 10^5 J + 6.33 x 10^5 J = ? is your answer if that's what you wanted to know
2007-11-01 09:21:10 · answer #3 · answered by Anonymous · 0⤊ 1⤋