A 25-kg child is placed on a swing with a chain length of 3.6-m Hler father pull

Question

A 25-kg child is placed on a swing with a chain length of 3.6-m Hler father pulls the swing back until it touches his chest, so that the chain is at 9.9-degrees to the vertical. He gives the swing a small push, so that it is travelling at 0.90 m/s. When the child reaches the apex of her motion, what is the new angle of the chain to the vertical? degrees Check How far back should the parent step so that the swing would just avoid hitting him when it returns (ignore air resistance and assume that the child does no work)? cm Check

Explanation / Answer

1) Let the new angle of the chain to the vertical =

The difference in vertical height from the start horizontal level = [3.6*cos9.9° - 3.6*cos]

conservation of energy:

loss in k.e = increase in potential energy

at the new apex position v = 0 => kinetic energy = 0

=> ½m(0.9)² = 3.6*mg[cos9.9° - cos]

=> ½(0.9)² = 3.6*9.8[cos9.9° - cos]

=> 0.405 = 35.28[cos9.9° - cos]

=> 0.405 = 34.75 - 35.28 cos

=> 35.28 cos = 34.75 - 0.405

=> cos = 0.97345

= 13.23 degrees --> Answer

2) horizontal distance difference between new apex position and start point = 3.6[sin13.23° - sin9.9°]

                                                                                                             = 0.205 m = 20.5 cm --> Answer

the parent should step back 21.35 cm to avoid the swing

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