Question

Q(25) A force F = 3 i + 6j+ 2 k and
Displacement S = - 4 i + 2) + 3 k then the work done is
a) 6J
b) 1 J
C) 12 J
d) ZeroJ​

Answer 1

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Answer 2

Answer:

Explanation:

Work done =force×displacement

Given data,force f=4i+j+3k N

Now displacement,

d=(3i+2j-6k)~(14i+13j+9k)

d=(14–3)i+(13–2)j+(9-[-6])k

d=11i+11j+15k

Now,W=f×d

W=(4i+j+3k)×(11i+11j+15k)

W=(11×4)+(1×11)+(3×15) °.° [i×i=1]

W=44+11+45

W=100 joules

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