physics questions (2 of them)?

Question

A parachutist with a mass of 50 kg jumps out of an airplane at an altitude of 1.00 X 10^3 m (1000m). After the parachute deploys the parachutist lands with a velocity of 5.00 m/s using the work-kinetic energy theorem, find the energy that was lost to air resistance during this jump (G= 9.81 m/s^2)

A. 49300 J
B. 98800 J
C. 198000 J
D. 489000 J
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A child riding a bike has a total mass of 40.0 kg the child approaches the top of a hill that is 10 m high and 100m long at 5.0m/s. If the force of friction between the bicycle and the hill is 20N, what is the childs velocity at the bottom of the hill? (Disregard air resistance g= 9.81 m/s^2)

A. 5.0m/s
B. 10.0 m/s
C. 11m/s
D. Child stops before reaching the bottom.


I've worked these again and again and i cannot come up with an answer (for the first one i got D but it was a stretch) anyways if anyone could post the answer and the work i would greatly X a million appreciate it. thanks.

I thought you would add the force of gravity and the force of friction to find the net force and then find acceleration and from there find the final velocity. But even if you do it like that you still get none of those answers so i'm thinking the answer might be D.

Answer 1

Potential Energy at top = mgh = 50x9.81x1000 = 490500 J
Kinetic energy at bottom = (mv^2)/2 = 50x5x5/2 = 625 J
Difference = 489875 J
All values have been given to 3 sig. figs. so our final answer will need to be the same.
Thus energy lost = 4.90x10^5 J
So D would be closest but they messed it up all the same.

Friction doesn't slow the motion of a bike, it converts it to rotation. We need to know the moment of inertia of the wheels of the bike and the bikes mass to answer this question properly so clearly the answers given in the multiple choice will be wrong. Anyway disregarding the loss of information the formulae are:
(KEchild+PEchild+KElinearbike+KErotationbike+PEbike)initial
=(KEchild+PEchild+KElinearbike+KErotationbike+PEbike)final
KEchildinitial = 500J
PEchildinitial-PEchildfinal = 3924J
therefore the childs final KE will be less than the difference = 3424J due to losses to increasing the rotation of the bike. The change in rotation of the wheel can be given by that the force of friction that acts is 20N over 100m or 2000J. The fact that this is a fixed force is completely unrealistic yet I'll continue anyway. This force acts to increase the rotational energy of the wheels. We have to assume the bike is massless (again unrealistic) which gives KEfinal = 3424J-2000J+500J = 1924J
therefore final velocity = 9.81m/s which again is none of the above.

Someone hasn't thought through these problems that well at all. You really shouldn't bother.