Question

A particle moves from position 3i^+2j^−6k^ to 14i^+13j^+9k^ due to uniform force of 4i^+j^+3k^N. If the displacement is in meters, then find the work done.

Answer 1

Given :

▪ Initial position = $\tt{3\hat{i}+2\hat{j}-6\hat{k}}$

▪ Final position = $\tt{14\hat{i}+13\hat{j}+9\hat{k}}$

▪ Force = $\tt{4\hat{i}+1\hat{j}+3\hat{k}}$N

To Find :

▪ Work done on the particle.

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Solution :

✒ Work done is defined as the dot product of force and displacement.

✒ It is a scalar quantity.

✒ It has only magnitude.

✒ SI unit : Nm = Joule(J)

✴ Displacement of particle :

$\rightarrow\sf\:\vec{d}=x_2-x_1\\ \\ \rightarrow\sf\:\vec{d}=(14\hat{i}+13\hat{j}+9\hat{k})-(3\hat{i}+2\hat{j}-6\hat{k})\\ \\ \rightarrow\sf\blue{\vec{d}=11\hat{i}+11\hat{j}+15\hat{k}}$

✴ Work done :

$\mapsto\sf\:W=\vec{F}\:{\tiny{\bullet}}\:\vec{d}\\ \\ \mapsto\sf\:W=(4\hat{i}+1\hat{j}+3\hat{k})(11\hat{i}+11\hat{j}+15\hat{k})\\ \\ \mapsto\sf\:W=44+11+45\\ \\ \mapsto\underline{\boxed{\bf{\gray{W=100\:J}}}}\:\orange{\bigstar}$

Answer 2

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$\Huge{\red{\underline{\textsf{Answer}}}}$

$\textbf{Given:–}$$\sf Force =4 \hat{\imath} + \hat{\jmath} + 3 \hat{k}N$

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$\textbf{Displacement}$

$\longrightarrow$ $\sf \vec{d} = (3 \hat{\imath} + 2 \hat{\jmath} - 6 \hat{k}) \sim (14 \hat{\imath} + 13 \hat{\jmath} + 19 \hat{k})$

$\longrightarrow$ $\sf \vec{d} = (14 - 3)i + (13 - 2)j + (9 - [-6])k$

$\orange\longrightarrow$ $\sf\orange{ \vec{d} = 11 \hat{\imath} + 11 \hat{\jmath} + 15 \hat{k}N}$

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Now, Work done = force × Displacement

$\leadsto$ $\sf W = (4 \hat{\imath} + \hat{\jmath} + 3 \hat{k}N) \times (11 \hat{\imath} + 11 \hat{\jmath} + 15 \hat{k}N)$

$\leadsto$ $\sf W = (11 \times 4) + (1 \times 11) + (3 \times 15)$

$\leadsto$ $\sf W = (44 + 11 + 45)$

$\purple\leadsto$$\large\underline\mathtt\purple{\fbox{W = 100 joules}}$

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