A force overrightarrow{F} = 6hat{i} + 2hat{j} | vedclass

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Question

(D) The work done $W$ by a force $\overrightarrow{F}$ during a displacement $\overrightarrow{s}$ is given by the dot product: $W = \overrightarrow{F}

Vedclass · Accepted Answer

$2$

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(D) The work done $W$ by a force $\overrightarrow{F}$ during a displacement $\overrightarrow{s}$ is given by the dot product: $W = \overrightarrow{F} \cdot \overrightarrow{s}$.Given $\overrightarrow{F} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ and $\overrightarrow{s} = 2\hat{i} - 3\hat{j} + x\hat{k}$.Since the work done is zero,we have $W = 0$.Substituting the vectors into the dot product formula:$(6\hat{i} + 2\hat{j} - 3\hat{k}) \cdot (2\hat{i} - 3\hat{j} + x\hat{k}) = 0$Calculating the dot product:$(6 	imes 2) + (2 	imes -3) + (-3 	imes x) = 0$$12 - 6 - 3x = 0$$6 - 3x = 0$$3x = 6$$x = 2$.

$A$ force $\overrightarrow{F} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ acts on a particle and produces a displacement of $\overrightarrow{s} = 2\hat{i} - 3\hat{j} + x\hat{k}$. If the work done is zero,the value of $x$ is

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