work done = F * distance
in this case the work done goes straight into increasing the kinetic energy
F * d = increase in Ke
same for both
one to one
the end.
A pickup truck (2268 kg) and a compact car (1100 kg) have the same momentum.
If the same horizontal net force were exerted on both vehicles, pushing them from rest over the same distance, what is the ratio of their final kinetic energies? (ratio: truck to car)
4 answers
That was intended to be a trick, confusing question, but you did not fall for it :)
By the way, of course they had the same momentum at rest.
Thanks man. I sorta figured out the hard way.
(mt)(at) = (mc)(ac)
at / ac = mc / mt
a = (vf² - vi²)/ 2∆x
at / ac = (vt² / 2∆x) / (vc² / 2∆x)
at / ac = vt² / vc²
mc / mt = vt² / vc² = 1100/2268
KE-t / KE-c = (mt)(vt²) / (mc)(ct²)
KE-t / KE-c = mt / mc * vt² / vc²
KE-t / KE-c = 2268/1100 * 1100/2268 = 1
sigh...
(mt)(at) = (mc)(ac)
at / ac = mc / mt
a = (vf² - vi²)/ 2∆x
at / ac = (vt² / 2∆x) / (vc² / 2∆x)
at / ac = vt² / vc²
mc / mt = vt² / vc² = 1100/2268
KE-t / KE-c = (mt)(vt²) / (mc)(ct²)
KE-t / KE-c = mt / mc * vt² / vc²
KE-t / KE-c = 2268/1100 * 1100/2268 = 1
sigh...