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A woman of mass 50kg swings on a light trapeze, i.e. a light seat suspended from a fixed point by a light rope.

Her centre of mass, G, moves on the arc of a circle of radius 9m and centre O. Initially the woman is at the point A, where OA is at 45 degrees to the vertical.

The first two parts of this question ask to calculate the gravitational potential energy lost as her centre of mass moves from A to B ( vertically below O), then assuming that there is no resistance to motion calculate her speed at B.

I approached part i using basic trigonometry and found that the loss in g.p.e. was approximately 1290J. For part ii, since $ 1/2 mv^2 = mgh$, i.e. $v^2=2gh$, it is simple to calculate that $ v = 7.19m/s $.

However I was unsure how to approach the second part:

A more refined model takes account of resistance by assuming that there is a force of constant magnitude 20N acting in the direction opposing motion. It is also assumed that the woman "pushes off" from A with a speed of 1m/s. Show that her speed B is approximately 6.86m/s.

Thanks in advance for any help.

Pie
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    As the force is assumed constant you can determine the energy lost to resistance as force x distance (=arc length), subtract this from the potential energy change, add her initial kinetic enrgy, and then detrmine her speed at B accordingly. – Tom Collinge Feb 20 '14 at 10:48
  • Ah I see now, thanks for the help – Pie Feb 20 '14 at 10:53
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    Remember, fundemental principle is conservation of energy in a closed system. Energy at start and end is the same, so initial potential + kinetic energy = final kinetic + potential + "dissapated" energy. The dissapated energy is in the form of heat, noise, air currents as a result of the work done against resistance. – Tom Collinge Feb 20 '14 at 11:38

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