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A particle is acted upon by constant forces $4 I+J-3 k$ and $3 I+J-k$ which displace it from a point $\hat{i}+2 \hat{j}+3 \hat{k}$ to the point $5 \hat{i}+4 \hat{j}+\hat{k}$. The work done in standard units by the forces is given by
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40
40
Work done by the forces $\vec{F}_1$ and $\vec{F}_2$ is $\left(\vec{F}_1+\vec{F}_2\right) \cdot \vec{d}$, where $\vec{d}$ is displacement According to question $\vec{F}_1+\vec{F}_2=(4 \hat{i}+\hat{j}-3 \hat{k})+(3 \hat{i}+\hat{j}-\hat{k})=7 \hat{i}+2 \hat{j}-4 \hat{k}$ and $\vec{d}=(5 \hat{i}+4 \hat{j}+\hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})=4 \hat{i}+2 \hat{j}-2 \hat{k}$. Hence $\left(\vec{F}_1+\vec{F}_2\right) \cdot \vec{d}$ is 40
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