Two constant forces F→ 1=2i^−3j^+3k^ (N) and F→ 2=i^+j^−2k^ (N) act on a body and displace it from the position r→ 1=i^+2j^−2k^ (m) to the position r→ 2=7i^+10j^+5k^ (m). What is the total work done in J.

W=F¯(r¯2−r¯1)=(3i^−2j^+k)(6i^+8j^+7k^)=18−16+7=9J

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Two constant forces ${\vec{F}}_{1} = 2 \hat{i} - 3 \hat{j} + 3 \hat{k}$ (N) and ${\vec{F}}_{2} = \hat{i} + \hat{j} - 2 \hat{k}$ (N) act on a body and displace it from the position ${\vec{r}}_{1} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ (m) to the position ${\vec{r}}_{2} = 7 \hat{i} + 10 \hat{j} + 5 \hat{k}$ (m). What is the total work done in J.

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answer is 9.

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$W = \bar{F} ({\bar{r}}_{2} - {\bar{r}}_{1})$

$= (3 \hat{i} - 2 \hat{j} + k) (6 \hat{i} + 8 \hat{j} + 7 \hat{k}) = 18 - 16 + 7 = 9 J$

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