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# the potential energy of theparticle moving along the x-axis is given by U(X)=8x^2+2x^4 where U is in the joule and x is in m.if total mechinal energy is 9j,then limits of motion are

148 Points
11 years ago

Dear jimmyu

mechanical energy of ay particle is equal to the sum of kinetic energy and potential energy

so                   1/2mV2 + U =9

or                   1/2mV2 + 8x2+2x4  =9

or                    1/2mV2 = 9 -  8x2-2x4

LHS is a positive value so RHs should also be positive value

9 -  8x2-2x   ≥0

or     -√17 -2√2        ≤ x2 ≤   √17- 2√2

but  0 ≤ x2

so                             0≤  x2 ≤   √17- 2√2

now   x2 ≤   √17- 2√2

x2 -(  √17- 2√2)  ≤0

x2 -{√(√17- 2√2)}2  ≤0

-√(√17- 2√2)   ≤ x   ≤       √(√17- 2√2)

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