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Tarzan swings on a 23.4 m long vine initially inclined at an angle of 32° from the vertical.

(a) What is his speed at the bottom of the swing if he starts from rest?

(b) What is his speed at the bottom of the swing if he starts with an initial speed of 2.28 m/s?

2006-11-07 14:46:34 · 4 answers · asked by tingerpoo 2 in Science & Mathematics Physics

4 answers

a. I agree with the answerers above about this problem involving the principle of conservation of energy.

At the point where Tarzan starts his downward swing, he is 3.56m above the bottom of the swing. We get this by solving for the length of the base (adjacent side) of the triangle whose hypotenuse equals the length of the vine, i.e. 23.4m. The length of the base is equal to 23.4cos32=19.84m. Deduct this from 23.4m and we get 3.56. Therefore,

PE=mg*3.56

and

KE=1/2mv^2

Equate the two based on our conservation of energy principle:

mg*3.56=1/2mv^2

(m cancels out)

v^2=69.77
v=8.35m/s

b)If he starts with an initial speed, then at that point where he starts, he has KE in addition to PE.

KE at start=1/2m*2.28^2

KE at bottom of swing=1/2mv^2

PE+KE at start=KE at bottom

mg*3.56+1/2m*2.28^2=1/2mv^2

(m cancels out)

34.88+2.59=1/2v^2
v^2=37.47*2
=74.94
v=8.66m/s

2006-11-08 06:41:56 · answer #1 · answered by tul b 3 · 0 0

Part A//////////

this is actually a pretty tricky question !!!

the tricky part is finding Tarzan's change in elevation. here goes:

imagine a triangle where one side is vertical with L=23.4 m, this is the position of the rope when Tarzan is at the lowest pt.

now imagine 2ND side of the triangle that's represented by the rope in it's initial position angled 32 degrees from the vertical

now draw a line joining the bottoms of the 2 ropes.

you should now have an isosceles triangle with the angle at the top = 32 degrees and the 2 side angles=(180 - 32)/2 = 74 degrees

let's find the length (L) of the base of the isosceles triangle above. we will be forced to use "sine law" because we don't have any "right angles".:

L/sin32=23.4/sin74 , L=12.9 m

now create yet another triangle using the 12.9 m length as the hypotenuse. I'm sure you can now see that the angle of the hypotenuse to the horizontal is (90 - 74)=16 degrees

finally, the change in elevation for Tarzan is:

h=12.9xsin16 = 3.5557

that was a pain in the "rump" finding h

anyway



his change PE = mgh, g=9.81, h=3.5557 and let m=M
PE=3.5557*9.81*M= 34.9M j

34.9M j must = KE at bottom of the swing

so KE = 0.5*M*v^2 = 34.9M, M's cancel out

V=(34.9/0.5)^0.5= 8.35 m/s

Part B////////////

his KE at the bottom will higher with an initial v=2.28 m/s

his total energy at the bottom now is:

34.9M j + 0.5*2.28^2 M = 37.5 M j

so to find his final V

0.5*M*(V_final)^2 = 37.5*M j, again the M's cancel out

V_final=(37.5/0.5)^0.5 = 8.66 m/s

2006-11-07 23:46:50 · answer #2 · answered by Mech_Eng 3 · 1 0

If he starts at rest, the angle and length tell how far he can fall vertically. He starts with potential energy and it's converted to kinetic energy at the bottom of the swing. If you set up the PE=KE equation (sorry, I don't have it off the top if my head) his mass isn't a consideration. That takes care of (a). For (b), his mass will also drop out. His initial energy will be the potential energy of the fall height PLUS the kinetic energy of moving at 2.28 m/s. That will be equal to his kinetic energy at the bottom of the swing. Use the equations in your book and move everything around to find the speed for both cases.

2006-11-07 23:10:20 · answer #3 · answered by Anonymous · 0 0

tarzan's last words were. "heyyyyy, who greased the grapevine"

2006-11-07 22:50:11 · answer #4 · answered by Anonymous · 0 3

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