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Problem number 80.
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Assuming the datum is at the level of the arc center.
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By applying the conservation of energy between 1 and 1, when theta equal 0, and 2, when theta equal 60 degrees.
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So, t1 plus v1 equals t2 plus v2.
00:27
So have mv1.
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V1 squared plus half k x1 squared plus m g h1 equal half mv2 squared plus m g h2 plus k x2 squared plus m g h 2 so it will equal to 0 plus 1 1 ,500 multiply 0 .1 squared minus 2, multiply .3, multiply 9 .81, multiply 1 .5, equal 0 .3b2 squared, plus 0 .0, minus 2, multiply 0 .3, multiply 9 .81, multiply 9 .81, 1 .5 cosine 60 degrees by simplifying the equation it will be 15 minus 8 .829 equal 0 .3 v2 squared minus 4 .414 so we will get that v2 is equal to 5 .94 meter per second and then from the equilibrium of the ball through this diagram this is t direction and this is n direction and this is the weight grafty direction the blue and this is 60 degrees and this is the force n okay, by applying the equilibrium equation, sigma f at n direction equal m, m, multiply v2 squared over rope...