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A 5.0 kilogram monkey hangs initially at rest from two vines, A and B. Each vine is 10 meters in length with negligible mass.

Vine A is on the left with an angle of 30 degrees from the horizontal from the monkey and B is at 60 degrees from the horizontal of the angle next to the monkey.

1. Determine the tension of B while it is at rest

2. Determine the speed of the monkey.

3. Determine the tension of vine B at the lowest point through the swing.

Here is what i got:

1. T=Wcos(theta)

=(5)(9.81)(cos 60)

=24.525N

2. PE(i)=KE(f)

mgh=.5(m)(v^2)

(5)(9.81)(10-10sin60)=(.5)(5)(v^2)

v=5.127

3. T=mg+F(centripital)

=(5)(9.81)+((5*5.127^2)/10m)

=62.192 N

Can anybody confirm this for me? I have been stuck on this problem for like 2 hours, Im confused if i calculated the tension in part 1 right or if i accidentlly did the tension for vine A instead. This is my worst part of physics, so any help would be appreciated!

Vine A is on the left with an angle of 30 degrees from the horizontal from the monkey and B is at 60 degrees from the horizontal of the angle next to the monkey.

1. Determine the tension of B while it is at rest

2. Determine the speed of the monkey.

3. Determine the tension of vine B at the lowest point through the swing.

Here is what i got:

1. T=Wcos(theta)

=(5)(9.81)(cos 60)

=24.525N

2. PE(i)=KE(f)

mgh=.5(m)(v^2)

(5)(9.81)(10-10sin60)=(.5)(5)(v^2)

v=5.127

3. T=mg+F(centripital)

=(5)(9.81)+((5*5.127^2)/10m)

=62.192 N

Can anybody confirm this for me? I have been stuck on this problem for like 2 hours, Im confused if i calculated the tension in part 1 right or if i accidentlly did the tension for vine A instead. This is my worst part of physics, so any help would be appreciated!

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