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Question
The work done by the force →F=2^i−3^j−2^k on a moving particle from (3,−2,5) to (1,−4,3) is
The work done by the force →F=2^i−3^j−2^k on a moving particle from (3,−2,5) to (1,−4,3) is
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Solution
The correct option is D 6
The displacement vector is
=i−4j+3k−(3i−2j+5k)
=−2i−2j−2k
Now W=→F.→d
=(2i−3j−2k).(−2i−2j−2k)
=−4+6+4
=6 joules
The displacement vector is
=i−4j+3k−(3i−2j+5k)
=−2i−2j−2k
Now W=→F.→d
=(2i−3j−2k).(−2i−2j−2k)
=−4+6+4
=6 joules
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