A force of 35N acts in the direction parallel to 2i + 3j + 6knand it displaces a body from (1 , 0, 3 )m to (3,4,1)m find the work done
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A force of 35N acts in the direction parallel to 2i + 3j + 6k. means, force along unit vector of 2i + 3j + 6k.
so, we can write F in vector form ,
F = 35{ (2i + 3j + 6k)/√(2² + 3² + 6²)}
= 35{ (2i + 3j + 6k)/7}
= 5(2i + 3j + 6k)
= 10i + 15j + 30k
A/C to question, particle displaces from (1, 0, 3) to (3, 4, 1) then, displacement vector = (3 - 1)i + (4 - 0)j + (1 - 3)k
= 2i + 4j - 2k
now, work is the dot product of force vector and displacement vector.
so, W = (10i + 15j + 30k).(2i + 4j - 2k)
=10 × 2 + 15 × 4 -30 × 2
= 20 + 60 - 60
= 20J
so, we can write F in vector form ,
F = 35{ (2i + 3j + 6k)/√(2² + 3² + 6²)}
= 35{ (2i + 3j + 6k)/7}
= 5(2i + 3j + 6k)
= 10i + 15j + 30k
A/C to question, particle displaces from (1, 0, 3) to (3, 4, 1) then, displacement vector = (3 - 1)i + (4 - 0)j + (1 - 3)k
= 2i + 4j - 2k
now, work is the dot product of force vector and displacement vector.
so, W = (10i + 15j + 30k).(2i + 4j - 2k)
=10 × 2 + 15 × 4 -30 × 2
= 20 + 60 - 60
= 20J
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