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Question
A constant force −→F1=(2^i−4^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 constant force −→F1=(2^i−4^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.
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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^i−4^j).(−2^i+^k)+(3^i+4^j).(6^i+6^j+2^k)
=−4+18+24=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^i−4^j).(−2^i+^k)+(3^i+4^j).(6^i+6^j+2^k)
=−4+18+24=38 J
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