A force →F=(5^i+3^j+2^k)N is applied over a particle which displaces it to a point having position vector→r=(2^i−^j)m from origin. The work done on the particle (in J) is
A
+10
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B
+7
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C
−7
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D
+13
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Solution
The correct option is B+7 Given constant force →F=(5^i+3^j+2^k)N Initial position →r1=(0^i+0^j+0^k)m, as starting from origin. Final position →r2=(2^i−^j+0^k)m
To find the displacement →s: →s=→r2−→r1= change in position →s=(2^i−^j+0^k)−(0^i+0^j+0^k) →s=(2^i−^j)m
Hence the work done , W=→F.→S W=(5^i+3^j+2^k).(2^i−^j)=10−3=7J