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Question
A particle is acted upon by constant forces 4^i+^j−3^k and 3^i+^j−^k which displace it from a point ^i+2^j+3^k to the point 5^i+4^j+^k. The work done is standard units by the forces is given by:
A particle is acted upon by constant forces 4^i+^j−3^k and 3^i+^j−^k which displace it from a point ^i+2^j+3^k to the point 5^i+4^j+^k. The work done is standard units by the forces is given by:
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Solution
The correct option is B 40
Total force =(4^i+^j−3^k)+(3^i+^j−^k)=7^i+2^j−4^k
The particle is displaced from A(^i+2^j+3^k) to B(5^i+4^j+^k)
Therefore displacement AB=(5^i+4^j+^k)−(^i+2^j+3^k)=4^i+2^j−2^k
Work done =F.AB=(7^i+2^j−4^k).(4^i+2^j−2^k)=28+4+8=40
Total force =(4^i+^j−3^k)+(3^i+^j−^k)=7^i+2^j−4^k
The particle is displaced from A(^i+2^j+3^k) to B(5^i+4^j+^k)
Therefore displacement AB=(5^i+4^j+^k)−(^i+2^j+3^k)=4^i+2^j−2^k
Work done =F.AB=(7^i+2^j−4^k).(4^i+2^j−2^k)=28+4+8=40
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