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jimmyu wazir Grade:
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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

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7 years ago

## Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
```										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 -  8x2-2x4    ≥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)
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin

```
7 years ago
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