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Question

Figure shows three forces applied to a block that moves leftward by 3 m over a smooth floor. The magnitudes of force are F1=5 N,F2=9 N and F3=3 N. The net work done on the block by the three forces will be


  1. 1.5 J
  2. 4.5 J
  3. 3 J
  4. 3.6 J  


Solution

The correct option is A 1.5 J

From the FBD of body and considering the rightwards as +ve x direction and upwards as +ve y direction ,
F=5^i+9cos60^i+9sin60^j3^j
=5^i+92^i+932^j3^j
=5^i+92^i+932^j3^j
F=12^i+(9323)^j
Displacement of the block S=3^i m
Work done by force,
W=FS=[12^i+(9323)^j](3^i)
W=1.5 J

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