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Question

Three forces F1=2^i4^j+5^k, F2=^i+3^j8^k and F3=6^i10^j+7^k are acting on a particle present on the positive x axis, 5 units away from the origin. The three forces combined displace the body to a point 6ˆi3ˆj+10ˆk. What is the net work done by the three forces ?

A
76 J
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B
70 J
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C
76 J
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D
70 J
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Solution

The correct option is B 70 J
Initial position of the particle = 5ˆi
Final position of the particle = 6ˆi3ˆj+10ˆk
Displacement = (6ˆi3ˆj+10ˆk)5ˆi=i3ˆj+10ˆk

Net force = F1+F2+F3=(2+16)ˆi+(4+310)ˆj+(58+7)ˆk=3ˆi11ˆj+4ˆk

Work done is given by,
W=F.sW=(3ˆi11ˆj+4ˆk).(ˆi3ˆj+10ˆk)W=3+33+40=70 J

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