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Question Number 58055 by Salomon M. last updated on 17/Apr/19
If the constant forces 2i−5j+6k and −i+2j−k act on a particle due to which it is displaced from a point A(4,−3,−2) to a point B(6, 1,−3), then the work done by the forces is
$$\mathrm{If}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{forces}\:\mathrm{2}\boldsymbol{\mathrm{i}}−\mathrm{5}\boldsymbol{\mathrm{j}}+\mathrm{6}\boldsymbol{\mathrm{k}}\:\mathrm{and} \\ $$$$−\boldsymbol{\mathrm{i}}+\mathrm{2}\boldsymbol{\mathrm{j}}−\boldsymbol{\mathrm{k}}\:\mathrm{act}\:\mathrm{on}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{due}\:\mathrm{to}\:\mathrm{which} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{displaced}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:{A}\left(\mathrm{4},−\mathrm{3},−\mathrm{2}\right) \\ $$$$\mathrm{to}\:\mathrm{a}\:\mathrm{point}\:{B}\left(\mathrm{6},\:\mathrm{1},−\mathrm{3}\right),\:\mathrm{then}\:\mathrm{the}\:\mathrm{work}\: \\ $$$$\mathrm{done}\:\mathrm{by}\:\mathrm{the}\:\mathrm{forces}\:\mathrm{is} \\ $$
Answered by Kunal12588 last updated on 17/Apr/19
Total Force F^→ =F_1 ^→ +F_2 ^→ F^→ =(2i−5j+6k)+(−i+2j−k) =i−3j+5k s_2 ^→ =6i+j−3k s_1 ^→ =4i−3j−2k s^→ =s_2 ^→ −s_1 ^→ =2i+4j−k W=F^→ ∙ s^→ =(i−3j+5k)∙(2i+4j−k) =2−12−5=−15
$${Total}\:{Force}\:\overset{\rightarrow} {{F}}=\overset{\rightarrow} {{F}}_{\mathrm{1}} +\overset{\rightarrow} {{F}}_{\mathrm{2}} \\ $$$$\overset{\rightarrow} {{F}}=\left(\mathrm{2}{i}−\mathrm{5}{j}+\mathrm{6}{k}\right)+\left(−{i}+\mathrm{2}{j}−{k}\right) \\ $$$$={i}−\mathrm{3}{j}+\mathrm{5}{k} \\ $$$$\overset{\rightarrow} {{s}}_{\mathrm{2}} =\mathrm{6}{i}+{j}−\mathrm{3}{k} \\ $$$$\overset{\rightarrow} {{s}}_{\mathrm{1}} =\mathrm{4}{i}−\mathrm{3}{j}−\mathrm{2}{k} \\ $$$$\overset{\rightarrow} {{s}}=\overset{\rightarrow} {{s}}_{\mathrm{2}} −\overset{\rightarrow} {{s}}_{\mathrm{1}} =\mathrm{2}{i}+\mathrm{4}{j}−{k} \\ $$$${W}=\overset{\rightarrow} {{F}}\:\centerdot\:\overset{\rightarrow} {{s}}=\left({i}−\mathrm{3}{j}+\mathrm{5}{k}\right)\centerdot\left(\mathrm{2}{i}+\mathrm{4}{j}−{k}\right) \\ $$$$=\mathrm{2}−\mathrm{12}−\mathrm{5}=−\mathrm{15} \\ $$
Commented by Kunal12588 last updated on 17/Apr/19
thanks sir
$${thanks}\:{sir} \\ $$
Commented by tanmay last updated on 17/Apr/19
kunal...(2i−5j+6k)+(−i+2j−k) =i−3j+5k s_2 =6i+j−3k s_1 =4i−3j−2k s_2 −s_1 =2i+4j−k W=(i−3j+5k).(2i+4j−k) =2−12−5 =−15
$${kunal}…\left(\mathrm{2}{i}−\mathrm{5}{j}+\mathrm{6}{k}\right)+\left(−{i}+\mathrm{2}{j}−{k}\right) \\ $$$$={i}−\mathrm{3}{j}+\mathrm{5}{k} \\ $$$${s}_{\mathrm{2}} =\mathrm{6}{i}+{j}−\mathrm{3}{k} \\ $$$${s}_{\mathrm{1}} =\mathrm{4}{i}−\mathrm{3}{j}−\mathrm{2}{k} \\ $$$${s}_{\mathrm{2}} −{s}_{\mathrm{1}} =\mathrm{2}{i}+\mathrm{4}{j}−{k} \\ $$$${W}=\left({i}−\mathrm{3}{j}+\mathrm{5}{k}\right).\left(\mathrm{2}{i}+\mathrm{4}{j}−{k}\right) \\ $$$$=\mathrm{2}−\mathrm{12}−\mathrm{5} \\ $$$$=−\mathrm{15} \\ $$

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