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A 27.0-kg child on a 1.00-m-long swing is released from rest when the ropes of the swing make an angle of 27.0° with the vertical. (a) Neglecting friction, find the child's speed at the lowest position. m/s (b) If the actual speed of the child at the lowest position is 1.30 m/s, what is the mechanical energy lost due to friction? J
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Added Fri, 07 Aug '15
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Let Potential energy at the lowest position be zero

At initial position the height of the swing from lowest position

= 1- cos 27 = 0.11 m

Kinectic Energy at lowest positon = Initial Potential Energy

\frac{1}{2}mv^{2} = mgh

v =\sqrt{2gh} = \sqrt{2 X 9.81 X 0.11} = 1.46 m/s

b) Loss in energy = \frac{1}{2}m\Delta v^{2} = 0.5 X 27 X (1.46^{2}-1.30^{2}) = 5.96 J

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