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Question-198860




Question Number 198860 by cherokeesay last updated on 25/Oct/23
Answered by mr W last updated on 25/Oct/23
D) is correct.  the force moved the block to a height  a sin θ, the work done is wa sin θ.  the force pulled the spring to a length  aθ, the work done is ((k(aθ)^2 )/2).  so the total work done by the force is  wa sin θ+((ka^2 θ^2 )/2).
$$\left.{D}\right)\:{is}\:{correct}. \\ $$$${the}\:{force}\:{moved}\:{the}\:{block}\:{to}\:{a}\:{height} \\ $$$${a}\:\mathrm{sin}\:\theta,\:{the}\:{work}\:{done}\:{is}\:{wa}\:\mathrm{sin}\:\theta. \\ $$$${the}\:{force}\:{pulled}\:{the}\:{spring}\:{to}\:{a}\:{length} \\ $$$${a}\theta,\:{the}\:{work}\:{done}\:{is}\:\frac{{k}\left({a}\theta\right)^{\mathrm{2}} }{\mathrm{2}}. \\ $$$${so}\:{the}\:{total}\:{work}\:{done}\:{by}\:{the}\:{force}\:{is} \\ $$$${wa}\:\mathrm{sin}\:\theta+\frac{{ka}^{\mathrm{2}} \theta^{\mathrm{2}} }{\mathrm{2}}. \\ $$
Commented by cherokeesay last updated on 25/Oct/23
Nice ! thank you master !
$${Nice}\:!\:{thank}\:{you}\:{master}\:! \\ $$

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