Question

Initial position of a particle is i^+5j^-k^ . When a force is applied to it. It's final position became 2i^- j^-k^. Find the work done by the force. If force is i^+j^+ k^ .
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Answer 1

Explanation:

u1 = i^+5j^-k^

u2 = 2i^-j^-k^

F = i^+j^+k^

we know that ,

Work done = F.S

S = u2 - u1 = i^-6j^+0k^

so , W = FxSx + FySy + FkSk ( in vector form )

W = (1x1)i^ + (1x -6)j^ + (1x0)k^

W = i^-6j^+0k^

Hence the value of work done is equall to

i^-6j^