A force F=51'+j'+4k^N is applied over a particle the displacement in this case is r=31'+j'_j^m.The work done on the particle is?
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Answer:
In this case, your particle starts from origin. Therefore, its final position r = 2i-j, is also its displacement. The work done is given by the following equation.
W = F.d
F = force vector
d = displacement vector
Notice how I used the dot to indicate scalar product.
Now,
W = (5i+3j).(2i-j)
On solving the above expression.
W = 5×2 - 3×1 = 10 - 3 = 7
Since the force is in Newtons and displacement is in metres, the work obtained will be in joules.
Therefore,
Your answer is
W = 7 joules
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