1.

When 1 N forceis applied increase in length ofthe springis 1 cm . Find elastic potential energystoredduringthis in it .

Answer»

`10 xx10^(-3)J `
`10^(-3)` J
`5XX10^(-3) J `
`20 xx10^(-3) J `

Solution :From ` F = - kx , k = F/x1/(10^(-2)) = 10^(2) `N/m
(neglectingnegativesign )
` U = 1/2 kx^(2) = 1/2 xx10^(2) XX (10^(-2))^(2)`
` = 5xx10^(-3) J ` .


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