Figure shows three forces applied to trunk that moves leftward by 3m over a smooth floor. The force magnitudes are F1 = 5N, F2 = 9N and F3 = 3N . The net work done on the trunk by the three forces
1.50 J
2.40 J
3.00 J
6.00 J
→ F = −5ˆi + 9 cos 60°ˆi +9 sin 60°ˆj − 3ˆj
= −5ˆi + 9/2ˆi +9√3/ 2 ˆj − 3ˆj
= −1/2 ˆi + (9√3/ 2 − 3)ˆ
→ s = −3ˆi.
W = → F . → s = [ ˆi. - 1 /2 +(9√3/ 2 − 3)ˆj]. (−3ˆi)
= 1.5 J
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