Question

The initial and final position vectors of a particle under a force F = 3i + 4j Newton are given
r = i + 2 j metre and ro = 4 i +6 j metre. Then the work done in Joule is:​

Answer 1

F = (3i + 4j) N

x1 = (i + 2j) m [initial position]

x2 = (4i + 5j) m  [final position]

We know that

Work done = F . s . cosФ                        [ here Ф=0 , cosФ=1]

s is displacement , then

s = x2 - x1

s = 4i + 5j - (i + 2j)

s = 4i + 5j - i -2j

s = 3i + 3j

Now,

Work done = F . s

W = (3i + 4j) . (3i + 3j) Nm

[∵ product of same vector components = 1 and that of different = 0]

W = i(3 * 3) + j(4 * 3) Nm

W = 9i + 12j Nm

W = (9i + 12j) J

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I hope you understand.