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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 \(............J\).

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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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