Work done by a force F on a body is W=F.S, where s is the displacement of body. Given that under a force F=(2i^+3j^+4k^) N a body is displaced from position vector r1=(2i^+3j^+k^) m to the position vector r2=(i^+j^+k^) m. Find the work done by this force.

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-8 J

4 J

-4 J

8 J

answer is A.

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Detailed Solution

The body is displaced from r1 to r2. Therefore, displacement of the body iss=r2-r1=(i^+j^+k^)-(2i^+3j^+k^)=(-i^-2j^)mNow, work done by the force is W=F·s=(2i^+3j^+4k^)·(-i^-2j^)=(2)(-1)+(3)(-2)+(4)(0)=-8 J

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