Question

A force of 35 N acts in the direction parallel to 2ˆi  3ˆj 6kˆ and it displaces a body from (1m, 0m,
3m) to (3m, 4m, 1m)
(a) Express the force vector (in unit vector form) (b) Find the work done

Answer 1

Answer:

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

Answer 2

Answer:

the answer is 20J

Explanation:

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