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Question

A constant force F1=(2^i4^j) N displaces a particle from (1,1,2) to (1,1,3) and then another force F2=(3^i+4^j) N displaces it from (1,1,3) to (5,5,5) (displacement being measured in metres). Find the total work done.

A
46 J
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B
38 J
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C
40 J
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D
30 J
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Solution

The correct option is B 38 J
Total work done =F1.d1+F2.d2
d1=(^i^j+3^k)(^i^j+2^k)
=2^i+0^j+^k
d1=2^i+^k
d2=(5^i+5^j+5^k)(^i^j+3^k)
=6^i+6^j+2^k
Total work done =(2^i4^j).(2^i+^k)+(3^i+4^j).(6^i+6^j+2^k)
=4+18+24=38 J

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