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sinequanon
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Homework Statement
This is a simple, yet rather convoluted problem, so please, bear with me.
An 80 Kg man on a sled is being chased by a lion down a 60 meter slope (μk = .13) with an incline of 30°. At the end of the slope, there is a 100 meter cliff. Also, there is a 20-meter-long rope at the end of the slope, which is tied to an overhanging tree branch. The rope is angled at 50° below the horizontal. The man grabs the rope and swings away to escape the lion. Let the point at the end of the slope = Point A, and let the point at which the rope is perpendicular to the ground = Point B.
Find:
1. The man's Potential Energy.
2. The Work of Friction.
3. The Velocity at Point A.
4. The Kinetic Energy at Point A.
5. The Potential Energy from Point A to Point B.
6. The total Kinetic Energy at Point B.
7. The Velocity at Point B.
8. The Tension in the Rope at Point B.
9. The angle of the rope when it reaches its highest point.
10. Assume that the lion is located at Point A when the man grabs the rope. Where does the lion land after falling off the cliff?
*Use 10 m/s² for gravity.
Homework Equations
Too many to name, but I'll try.
PE = mgy
W = μk*n*d
KE = (1/2)m(v²)
F(centripetal) = (mv²)/r
Tension = F(centripetal) + ω
And other basic equations. I'm just having a difficult time factoring each of them in.
The Attempt at a Solution
Well, firstly, finding the initial PE is fairly simple: PE = (80 Kg)(10 m/s²)(60sin30°) = 24000 J (#1).
Then, I assume I would use the equation W(Friction) = μk*n*d, so W(Friction) = (.13)(800cos30°)(60m) = 692.82 J (#2).
I would then procede to take the difference between the two, so PE - W(Friction) = 24000 - 692.82 = 23307.18 J, which would be the Kinetic Energy at Point B (#4).
By then making that equal to the equation for KE, I would would use 23307.18 = (1/2)(80)(v²), and solving for v, which would equal 24.14 m/s (#3).
Now, the fifth part is worded ambiguously to me. So, from here, I'm pretty much lost. Knowing my teacher, I doubt he is asking for a representation of the PE for all points between A and B. I suspect it is merely supposed to mean what the PE is at Point A. However, if that is true, I am unsure as to how to go about factoring the initial velocity into the equation (Our curriculum is all over the map, and I can't seem to find it in my textbook). I also have an inkling that I made a mistake in the first part of the equation, so I would appreciate a double-check.
Please, if you could, help me. Thank you
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