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Question

Three constant forces are applied on a body that 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 the origin, then total work done by the forces is

A
20 N
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B
65 N
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C
5 N
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D
20 N
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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^j0^j=5^j
Work done W=F.ds=(3^i+4^j+6^k).(5^j)=20 N

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