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Question

A force of magnitude 18 N acts on a particle in the direction of a vector A=2^i4^j+4^k, which results in the change of position of the particle from (3,3,5) m to (2,1,4) m. The work done by the force is (in Joule )

A
66
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B
30
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C
20
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D
40
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Solution

The correct option is B 30
Given, a force is acting in the direction of the vector A=2^i4^j+4^k,
So, we have
F=|F|.^F=|F|.^A=|F|.A|A|
F=(18)⎜ ⎜2^i4^j+4^k22+(4)2+42⎟ ⎟=6^i12^j+12^k
Now, we have position vectors as
r1=3^i+3^j+5^k and r2=2^i^j+4^k
So, the displacement vector is given as
S=r2r1=^i4^j^k
Thus, the work done by the force is given as
W=F.S=(6^i12^j+12^k).(^i4^j^k)
W=6+4812=30 Joule

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