A mass hangs on the end of a massless rope. The pendulum is held horizontal and released from rest. When the mass reaches the bottom of its path it is moving at a speed v = 2.5 m/s and the tension in the rope is T = 15.9 N.

5) Return to the original mass. What is the tension in the string at the same vertical height as the peg (directly to the right of the peg)?

Question

A mass hangs on the end of a massless rope. The pendulum is held horizontal and released from rest. When the mass reaches the bottom of its path it is moving at a speed v = 2.5 m/s and the tension in the rope is T = 15.9 N.

5) Return to the original mass. What is the tension in the string at the same vertical height as the peg (directly to the right of the peg)?

anonymous · Answer

I have 1-4 solved correctly.
mass= .5403
length of rope=.3185
max tension= 33.84
speed the mass right before it hits peg=2.24

anonymous · Answer

I think the equation I want to use is T= F-Mg

noelgreco · Answer

There is no mention of a peg in the original question.

Can you provide more information?  A drawing is always a big help.

anonymous · Answer

attachment

anonymous · Answer

Additional info
Now a peg is placed 4/5 of the way down the pendulum’s path so that when the mass falls to its vertical position it hits and wraps around the peg. As it wraps around the peg and attains its maximum height it ends a distance of 3/5 L below its starting point (or 2/5 L from its lowest point).
How fast is the mass moving at the top of its new path (directly above the peg)?

anonymous · Answer

nite here a good app http://www.mathopenref.com/calculator.html