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A 1 kg block subjected to two simultaneous forces (2 i+3 j+4 k ) N and (3 i- j-2 k ) N is moved a distance of 25 m along (3 i-4 j ) direction. The work done in this process is J.
35
Net force: F= F1+ F2=(2+3) i+(3-1) j+(4-2) k=5 i+2 j+2 k N
Displacement of 25 m along (3 i-4 j):
d=25 3 i-4 j 32+42=15 i-20 j m
W= F d=5 15+2 (-20)+2 0
⇒ W=75-40=35 J
Answer: 35
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