Question

When the load on a wire is slowly increased from `3kgwt` to `5 kg wt`, the elongation increases from `0.61` to `1.02 mm`. The work done during the extension of wire is

Answer 1

Here, `F_(1)=3kg f=3 xx 9.8 N, Delta l_(1) = 0.61 xx 10^(-3)m`
`F_(2)=5kg f=5 xx 9.8 N, Delta l_(2) = 1.02 xx 10^(-3)m`
Work done, `W=1.2 F_(2) Delta l_(2) - 1/2 F_(1)Delta l_(1)`
`=1/2 [(5 xx 9.8) xx (1.02 xx 10^(-3))`
`-(3 xx 9.8) xx (0.61 xx 10^(-3))]`
`=16.023 xx 10^(-3)J`.