Question

A force $(10\hat{i}-3\hat{j}+6\hat{k})$ newton acts on a body of mass $100g$ and displaces it from $(6\hat{i}-5\hat{j}-3\hat{k})$ metre to $(10\hat{i}-2\hat{j}+7\hat{k})m$
The work done is

• A. $21J$
• B. $361J$
• C. $121J$
• D. $1000J$

Answer 1

Answer: (C)

$121J$

Force $\overrightarrow{F}=10\hat{i}-3\hat{j}+3\hat{k}$

Mass $M=100g$

Displacement $(6\hat{i}-5\hat{j}-3\hat{k})(10\hat{i}-2\hat{j}+7\hat{k})$

Displacement $\overrightarrow{d}=4\hat{i}-7\hat{j}+10\hat{k}$

Work $=$ Force $.$ displacement

$=\overrightarrow{F}.\overrightarrow{d}=(10\hat{i}-3\hat{j}+6\hat{k})(4\hat{i}-7\hat{j}+6\hat{k})=(10\times 4)+(7\times 3)+(10\times 6)=121J$