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For examples 15, find the total energy of the block at 0.05 m from the meanposition. Also,show that it issame as the P.E. Hint : Total energy =KE +PE at maximum displacement , KE=0 ,hence, total energy =PE |
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Answer» Solution :TOTAL energy of the BLOCK at `x=0.05 m` `= KE +PE = ( -.375 + 0.125 ) J` `=0.5 J` We also know that at maximum displacement , `KE = 0` and hence, the total energy of the system `= P.E. ` The total energy of the system `=(1)/(2) xx 100 xx0.1 xx 0.1 J` `=0.5 J` which ISSAME as the sum of the TWO energies ata displacement of 0.05 m . This is in CONFORMITY with the principles of conservation of energy. |
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