Question

A 28.0 kg child plays on a swing having support ropes that are 1.90 m long. A friend pulls her back until the ropes are at angle ϕ = 43.0 ∘ 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?

          A 28.0 kg child plays on a swing having support ropes that are
1.90 m long. A friend pulls her back until the ropes are at angle ϕ
= 43.0 ∘ 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?
        
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Added by Melissa C.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A 28.0 kg child plays on a swing having support ropes that are 1.90 m long. A friend pulls her back until the ropes are at angle ϕ = 43.0 ∘ 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?
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Transcript

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00:01 So in this question we're thinking about a 28 kilogram child playing on a swing, who is pulled back by an angle of phi, which is 43 .0 degrees, on ropes that are l equals 1 .90 metres long.
00:23 So what's the potential energy for the child just before she's released? u, the potential energy, is m, g, h.
00:32 So that's m g times the sign of phi.
00:45 Sorry, that's not right, it's 1 minus the cos of phi.
00:54 So we can work that out with a calculator.
00:56 We've got 28 times g, which is 9 .81 meters per second squared, times 1 minus the cosine of 43 degrees.
01:06 And that gives us 73 .79 joules.
01:14 So then how fast will she be moving? so that's part a.
01:18 Part b is how fast is she moving at the bottom of the swing? well, we're going to convert this potential energy into kinetic energy...
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