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For examples 15, find the potential energy of the block at0.05 m from the mean position Hint : P.E. = (1)/(2) kx^(2) |
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Answer» SOLUTION :P.E. of the block `= (1)/(2) kx^(2)` `= (1)/(2) XX 100 xx 0.05 ^(2)` `= 0.125 J` |
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