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A force magnitude $5$ units acting parallel to $\overrightarrow{2i}-\overrightarrow{2j}+ \overrightarrow{k}$ displaces the point of application from $(1,2,3)$to $(5,3,7) $ find the work done.

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  • The work done by a force $ \overrightarrow F$ in displacing a particle through $ \overrightarrow d$ is $ \overrightarrow F.\overrightarrow d = |\overrightarrow F| $ ( displacement in the direction of $ \overrightarrow F$)
  • The unit vector in the direction of $ \overrightarrow a$ is given by $ \overrightarrow a = \large\frac{\overrightarrow a}{|\overrightarrow a|}$
The particle is displaced from $ A(1,2,3) [ \overrightarrow {OA} = \overrightarrow i+2 \overrightarrow j+3 \overrightarrow k]$ to $B(5,3,7) [ \overrightarrow {OB}=5 \overrightarrow i+3 \overrightarrow j+7 \overrightarrow k]$
The displacement is $ \overrightarrow d = \overrightarrow {OB}- \overrightarrow {OA} = 4 \overrightarrow i+ \overrightarrow j+4 \overrightarrow k$
$ \overrightarrow F=5 \overrightarrow a$ where $ \overrightarrow a$ is the unit vector in the direction $ \overrightarrow a=2 \overrightarrow i-2 \overrightarrow j+ \overrightarrow k.\: | \overrightarrow a| = \sqrt{4+4+1}=3$
$ \therefore \overrightarrow a = \large\frac{1}{3}$$ ( 2 \overrightarrow i-2 \overrightarrow j+ \overrightarrow k) \: and \: \overrightarrow F = \large\frac{5}{3}$$ ( 2 \overrightarrow i-2 \overrightarrow j+ \overrightarrow k)$
$ \therefore $ work done $ w = \overrightarrow f. \overrightarrow d=\large\frac{5}{3}$$ (2 \overrightarrow i-2 \overrightarrow j+ \overrightarrow k).(4 \overrightarrow i+ \overrightarrow j+4 \overrightarrow k)$
$ = \large\frac{5}{3} $$( (2)(4)+(-2)(1)+(1)(4))$
$ = \large\frac{5}{3}$$ [ 8-2+4]= \large\frac{50}{3}$ units
answered Jun 4, 2013 by thanvigandhi_1
edited Jul 18, 2013 by balaji.thirumalai

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