Question

Under a force $10\hat{i}-3\hat{j}+8\hat{k}N$ a body of mass $5kg$ moves from position $(6\hat{i}+5\hat{j}-3\hat{k})m$ to $(10\hat{i}-2\hat{j}+7\hat{k})m$. The work done by the force is:

Answer 1

$W=\overrightarrow{F}.\overrightarrow{d}=\left(\begin{array}{c}10\hat{i}-3\hat{j}+8\hat{k}\end{array}\right).\left(\begin{array}{c}10\hat{i}-2\hat{j}+7\hat{k}-6\hat{i}-5\hat{j}+3\hat{k}\end{array}\right)$

$=\left(\begin{array}{c}10\hat{i}-3\hat{j}+8\hat{k}\end{array}\right).\left(\begin{array}{c}4\hat{i}-7\hat{j}+10\hat{k}\end{array}\right)$

$=40+21+80=141J$