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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A body moves a distance 10m along a straight line under the action of force $5N.$ If the work done is $25J$, the angle which the direction of motion of body is

\[(a)\;0^{\circ} \quad (b)\;60^{\circ} \quad (c)\; 30^{\circ} \quad (d)\;90^{\circ}\]

Can you answer this question?
 
 

1 Answer

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$W=F.S$
$\quad=FS \cos \theta$
$25=5 \times 10 \cos \theta$
$\cos \theta =\large\frac{25}{50}=\frac{1}{2}$
$\theta= 60^{\circ}$
Hence  B is the correct answer.
answered Aug 6, 2013 by meena.p
edited Apr 2, 2014 by balaji.thirumalai
 

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