Three constant forces are applied on a body that are given as −→F1=3^iN,−→F2=4^jN,−→F3=6^kN.
If these forces displace the body by 5m in negative y-axis direction from the origin, then total work done by the forces is
A
20N
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B
65N
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C
−5N
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D
−20N
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Solution
The correct option is D−20N −−→Fnet=→F1+→F2+→F3 →Fnet=(3^i+4^j+6^k)N
For displacement vector:
Initial position =0
Final position=−5^j →ds=−5^j−0^j=−5^j
Work done W=→F.→ds=(3^i+4^j+6^k).(−5^j)=−20N