A 25kg child plays on a swing having support ropes that are 2.20m long. A friend pulls her back until the

Question

ropes are 42 degrees from the vertical and releases her from rest.a)what is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing? b)how fast will she be moving at the bottom of the swing? c)How much work does the tension in the ropes do as the child swings from the initial position to the bottom?

Answer 1

a) PE = mgh

h is the vertical difference of when the swing is at rest (2.2 m) and when the child is pulled back.

Find the height when the child is pulled back:

cos 42 = y/22

y = 2.20 cos 42 = 1.63 m

Therefore, the difference (2.20 m - 1.63 m = .57 m) is h

PE = (25 kg)(9.8 m/s²)(.57 m) = 138.44 J

b) At the bottom, all the PE will be transferred into KE

KE = 1/2mv²

mv² = 2KE

v² = 2KE/m

v = √(2KE/m)

v = √(2*138.44 J/25 kg)

v = √11.08 J/kg

v = 3.33 m/s

c) Work = Fd

F = mg = 25 kg * 9.8 m/s² = 245 N

d is the rope exends. The rope doesn't extend so d = 0

245 * 0 =

Work = 0 J

Answer 2

A) POTENTIAL ENERGY "U"=mgh and h=2.2-(2.2(cos42))=.565m

                                       U=25(9.81)(.565) = 138.57 kg m^2/s^2 or

                                       U= 138.57 joules,

at the bottom of the swing the all the potential energy has become kenetic energy so potential energy will be zero and kinetic energy "T" will be 138.57 joules.

B) Kinetic Energy "T"=m(v^2/2)  

                                        138.57=25(v^2/2)

                                         v=3.33 meters/second

C) WORK "W"=Usub2-Usub1 plus Tsub2-Tsub1

                 W=0-138.57 plus138.57-0

                 W=0

or work is force times distance, and since the distance of th ropes remained constant the change in distance is zero, therefore work is zero.