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Home  >>  TN XII Math  >>  Vector Algebra
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The constant forces $\overrightarrow{2i}-\overrightarrow{5j}+ \overrightarrow{6k}, -\overrightarrow{i}+\overrightarrow{2j}-\overrightarrow{k},$and $\overrightarrow{2i}+\overrightarrow{7j} $ act on a particle which is displaced from position $\overrightarrow{4i}-\overrightarrow{3j}-\overrightarrow{2k}$ to position $\overrightarrow{6i}+\overrightarrow{j}-\overrightarrow{3k}$. find the work done.

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  • If a number of forces $ \overrightarrow F_1, \overrightarrow F_2, \overrightarrow F_3, ....$ act on a particle and displace it through $ \overrightarrow d$, then the work done by the force is $ ( \overrightarrow F_1+\overrightarrow F_2+\overrightarrow F_3...).\overrightarrow d$
Here the initial position of the particle is given by $ \overrightarrow {OA}=4\overrightarrow i-3\overrightarrow j-2\overrightarrow k$ and the final position by $ \overrightarrow {OB}=6\overrightarrow i+\overrightarrow j-3\overrightarrow k$.
The displacement $ \overrightarrow d=\overrightarrow {AB}=\overrightarrow {OB}-\overrightarrow {OA}=2\overrightarrow i+4\overrightarrow j-\overrightarrow k$
The forces acting on the particle are
$ \overrightarrow F_1=2\overrightarrow i-5\overrightarrow j+6\overrightarrow k,\: \overrightarrow f_2 = -\overrightarrow i+2\overrightarrow j-\overrightarrow k, \: \overrightarrow F_3=2\overrightarrow i+7\overrightarrow j$
The resultant force is $ \overrightarrow F=\overrightarrow F_1+\overrightarrow F_2+\overrightarrow F_3 = 3\overrightarrow i+4\overrightarrow j+5\overrightarrow k$
The work done by the forces is $ \overrightarrow f.\overrightarrow d = (3\overrightarrow i+4\overrightarrow j+5\overrightarrow k).(2\overrightarrow i+4\overrightarrow j-\overrightarrow k)$
$ (3)(2)+(4)(4)+(5)(-1)$
$ = 6+16-5=17$ units.

 

answered Jun 20, 2013 by thanvigandhi_1
 

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