JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2023
- Easy
- Numerical
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 \(............J\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(40\)
Step-by-step Solution
Detailed explanation
\(W=\vec{F} \cdot\left(\vec{r}_f-\vec{r}_{ i }\right)\)
\(=(5 \hat{i}+2 \hat{j}+7 \hat{k}) \cdot((5 \hat{i}-2 \hat{j}+\hat{k})-(2 \hat{i}+3 \hat{j}-4 \hat{k}))\)
\(W=40\,J\)
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