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In this question take

A particle of mass 0.5 kg is suspended from a fixed point O by a light elastic string of natural length 1.5 m and modulus of elasticity 12 N. The particle is released from rest at O and next comes to rest at the point A. When the particle is at the point A, the extension in the string is x

metres. Show, using energy considerations, that 8x2 -10x-15=0

Assuming that after it says "In this question take" it says g=10m/s^2 and you mean k = 12N/m (12N is a force), then:Elastic potential energy = 0.5kx^2 = 0.5×12×x^2 = 6x^2Gravitational potential energy = mgh = 0.5×10×(1.5 + x) = 5(1.5 + x) = 5x + 7.5Kinetic energy starts at zero and finishes at zeroDue to the conservation of energy, the sum of all energy changes = 0Gain in elastic potential energy − loss of gravitational potential energy = 06x^2 − (5x + 7.5) = 06x^2 − 5x − 7.5 = 012x^2 − 10x − 15 = 0Which is not the answer, but this is correct using the numbers given. I hope you can follow the method and work it out.
Sam P.
05 May 2016
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