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A force $m \hat{i}-3 \hat{j}+\hat{k}$ acts on a point and so the point moves
from $(20,3 \mathrm{~m}, 0)$ to $(0,0,7)$. If the work done by the force is $-48$ unit, what is the value of $\mathrm{m}$ ?
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from $(20,3 \mathrm{~m}, 0)$ to $(0,0,7)$. If the work done by the force is $-48$ unit, what is the value of $\mathrm{m}$ ?
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Verified Answer
The correct answer is:
5
Force, $\overrightarrow{\mathrm{F}}=\mathrm{m} \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
Due to this force, point moves from $\mathrm{A}(20,3 \mathrm{~m}, 0)$ to $\mathrm{B}(0,0,$,7). So, the displacement vector $\overrightarrow{\mathrm{AB}}$ is given by $\overrightarrow{\mathrm{AB}}=-20 \hat{\mathrm{i}}-3 \mathrm{~m} \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$
Work done $=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{AB}}$
$=(\mathrm{m} \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot(-20 \hat{\mathrm{i}}-3 \mathrm{~m} \hat{\mathrm{j}}+7 \hat{\mathrm{k}})$
$=(-20 \mathrm{~m}+9 \mathrm{~m}+7)$ unit
But work done $=-48$ unit, as given $\Rightarrow-11 \mathrm{~m}+7=-48$
$\Rightarrow-11 \mathrm{~m}=-55$
$\Rightarrow \mathrm{m}=5$
Due to this force, point moves from $\mathrm{A}(20,3 \mathrm{~m}, 0)$ to $\mathrm{B}(0,0,$,7). So, the displacement vector $\overrightarrow{\mathrm{AB}}$ is given by $\overrightarrow{\mathrm{AB}}=-20 \hat{\mathrm{i}}-3 \mathrm{~m} \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$
Work done $=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{AB}}$
$=(\mathrm{m} \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot(-20 \hat{\mathrm{i}}-3 \mathrm{~m} \hat{\mathrm{j}}+7 \hat{\mathrm{k}})$
$=(-20 \mathrm{~m}+9 \mathrm{~m}+7)$ unit
But work done $=-48$ unit, as given $\Rightarrow-11 \mathrm{~m}+7=-48$
$\Rightarrow-11 \mathrm{~m}=-55$
$\Rightarrow \mathrm{m}=5$
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