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Question Number 58056 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 tanmay last updated on 17/Apr/19

OA^→ =4i−3j−2k OB^→ =6i+j−3k displacement=AB^→ =OB^→ −OA^→ AB^→ =(6i+j−3k)−(4i−3j−2k) AB^→ =2i+4j−k F_1 ^→ +F_2 ^→ =(2i−5j+6k)+(−i+2j−k) F_1 ^→ +F_2 ^→ =i−3j+5k W=(F_1 ^→ +F_2 ^→ ).(AB^→ ) =(i−3j+5k).(2i+4j−k) =2×1+(−3)×4+5×(−1) =2−12−5 =−15 unit

$${O}\overset{\rightarrow} {{A}}=\mathrm{4}{i}−\mathrm{3}{j}−\mathrm{2}{k} \\ $$$${O}\overset{\rightarrow} {{B}}=\mathrm{6}{i}+{j}−\mathrm{3}{k} \\ $$$${displacement}={A}\overset{\rightarrow} {{B}}={O}\overset{\rightarrow} {{B}}−{O}\overset{\rightarrow} {{A}} \\ $$$${A}\overset{\rightarrow} {{B}}=\left(\mathrm{6}{i}+{j}−\mathrm{3}{k}\right)−\left(\mathrm{4}{i}−\mathrm{3}{j}−\mathrm{2}{k}\right) \\ $$$${A}\overset{\rightarrow} {{B}}=\mathrm{2}{i}+\mathrm{4}{j}−{k} \\ $$$$\overset{\rightarrow} {{F}}_{\mathrm{1}} +\overset{\rightarrow} {{F}}_{\mathrm{2}} \:=\left(\mathrm{2}{i}−\mathrm{5}{j}+\mathrm{6}{k}\right)+\left(−{i}+\mathrm{2}{j}−{k}\right) \\ $$$$\overset{\rightarrow} {{F}}_{\mathrm{1}} +\overset{\rightarrow} {{F}}_{\mathrm{2}} ={i}−\mathrm{3}{j}+\mathrm{5}{k} \\ $$$${W}=\left(\overset{\rightarrow} {{F}}_{\mathrm{1}} +\overset{\rightarrow} {{F}}_{\mathrm{2}} \right).\left({A}\overset{\rightarrow} {{B}}\right) \\ $$$$\:\:\:=\left({i}−\mathrm{3}{j}+\mathrm{5}{k}\right).\left(\mathrm{2}{i}+\mathrm{4}{j}−{k}\right) \\ $$$$\:\:\:=\mathrm{2}×\mathrm{1}+\left(−\mathrm{3}\right)×\mathrm{4}+\mathrm{5}×\left(−\mathrm{1}\right) \\ $$$$=\mathrm{2}−\mathrm{12}−\mathrm{5} \\ $$$$=−\mathrm{15}\:{unit} \\ $$

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