00:01
All right, so let's say we have a skier being towed up an incline at 40 degrees, and i'm just going to model the skier as a box.
00:09
But the tow rope has some kind of tension in it, and we're told that after traveling a distance of 2 .18 meters, the rope slips out of his hands.
00:22
And his speed, we're told, at the top of the ramp is, we'll call it's vf5 meters per second.
00:30
We want to know if his speed at the bottom ramp is 15 meters per second, what is the speed, or what is the tension in the rope? so basically, the change in kinetic energy is going to be equal to the work done by friction, plus the work done by gravity, plus the work done by the tension in the rope, on the point till the rope slips out.
00:59
So if we write this out, this is going to be like minus mu, m g cosine theta that's our normal force times delta s minus m g sine theta times delta s plus the work done by the tension which is just positive t delta s so this should give us our change in kinetic energy up until the point where the rope slips and this is just the change in kinetic energy imparted while he's in contact with the rope after he lets go the rope his changing kinetic energy, we'll call this delta k2, so the second part of the problem.
01:39
It's just going to be the work done by friction, so muk times m g times the cosine of theta times 10 meters minus delta s, that's the remaining distance he travels, minus m g sine theta times 10 meters minus delta s.
02:02
And so we can, if we add these two together, so this is going to be like one half times the final velocity of stage one, or sorry, the final velocity total minus the final velocity of stage one.
02:16
So we'll just call that v1 squared.
02:19
That should be equal to this...