Work Power Energy Question 7
Question 7 - 01 February - Shift 1
A small particle moves to position $5 \hat{i}-2 \hat{j}+\hat{k}$ from its initial position $2 \hat{i}+3 \hat{j}-4 \hat{k}$ under the action of force $5 \hat{i}+2 \hat{j}+7 \hat{k} N$. The value of work done will be $J$.
Show Answer
Answer: (40)
Solution:
Formula: Work Done By Constant Force
$W=\vec{F} \cdot( \vec{r} _f- \vec{r} _i)$
$=(5 \hat{i}+2 \hat{j}+7 \hat{k}) \cdot((5 \hat{i}-2 \hat{j}+\hat{k})-(2 \hat{i}+3 \hat{j}-4 \hat{k}))$
$W=40 J$
