Two forces each of magnitude $10$ N act simultaneously on a body with their directions inclined to each other at an angle of $120^{0}$ and displaces the body over $10$ m along the bisector of the angle between the two forces. Then the work done by each force is:

5 J

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1 J

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50 J

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100 J

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Solution

The correct option is D 50 J
Work done by each force $= \int \to F . d \to s$

F and s should be in the same direction so the component of F is taken
$= 10.10 cos 60^{0}$
$= 100 cos 60$
$= 50 J$

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Two forces each of magnitude 10 N act simultaneously on a body with their directions inclined to each other at an angle of 1200 and displaces the body over 10 m along the bisector of the angle between the two forces. Then the work done by each force is:

The correct option is D 50 JWork done by each force =∫→F.d→sF and s should be in the same direction so the component of F is taken=10.10cos600=100cos60=50 J

Two forces each of magnitude $10$ N act simultaneously on a body with their directions inclined to each other at an angle of $120^{0}$ and displaces the body over $10$ m along the bisector of the angle between the two forces. Then the work done by each force is:

The correct option is D 50 J
Work done by each force $= \int \to F . d \to s$

F and s should be in the same direction so the component of F is taken
$= 10.10 cos 60^{0}$
$= 100 cos 60$
$= 50 J$