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Question
Three constant forces are applied on a body are given as −→F1=3^i N,−→F2=4^j N,−→F3=6^k N.
If these forces displace the body by 5 m in negative y-axis direction from origin. Then total work done by the forces is
Three constant forces are applied on a body are given as −→F1=3^i N,−→F2=4^j N,−→F3=6^k N.
If these forces displace the body by 5 m in negative y-axis direction from origin. Then total work done by the forces is
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Solution
The correct option is D −20 N
−−→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)=−20 N
−−→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)=−20 N
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