A 28.0 kg child plays on a swing having support ropes that are 1.80 m long. A fr

Question

A 28.0 kg child plays on a swing having support ropes that are 1.80 m long. A friend pulls her back until the ropes are 43.0 from the vertical and releases her from rest.

part 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?

U=      J

part b- How fast will she be moving at the bottom of the swing?

v=                 m/s

part c- How much work does the tension in the ropes do as the child swings from the initial position to the bottom?

W=                     J

Explanation / Answer

GIVEN: m=28kg,length=1.8m, angle=43deg

a) PE = mgh

h is the vertical difference of when the swing is at rest (1.8 m) and when the child is pulled back.

Find the height when the child is pulled back:
cos 43 = y/1.8
y = 1.8cos 43 = 1.31 m

Therefore, the difference (1.8 m - 1.31 m = 0.48 m) is h
PE = (28 kg)(9.8 m/s²)(0.48 m) = 131.712 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*131.712 J/228 kg)
v = 3.06 m/s

c) Work = Fd
F = mg = 28 kg * 9.8 m/s² = 274.4 N
d is the rope exends. The rope doesn't extend so d = 0
274.4 * 0 =
Work = 0 J

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