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# A force of magnitude 18 N acts on a particle in the direction of a vector →A=2^i−4^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 30Given, a force is acting in the direction of the vector →A=2^i−4^j+4^k, So, we have →F=|→F|.^F=|→F|.^A=|→F|.→A|→A| ⇒ →F=(18)⎛⎜ ⎜⎝2^i−4^j+4^k√22+(−4)2+42⎞⎟ ⎟⎠=6^i−12^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=→r2−→r1=−^i−4^j−^k Thus, the work done by the force is given as W=→F.→S=(6^i−12^j+12^k).(−^i−4^j−^k) ⇒ W=−6+48−12=30 Joule  Suggest Corrections  0      Similar questions  Explore more