A force F=(5i^-3i^+2k^)N moves a particle from r1=(2i^+7j^+4k^)m to (5i^+2j^+8k^)m. The work done by the force is
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Given
F= 5i-3j+2k
now, distance(x) =r2-r1
X= (5i+2j+8k)-(2i+7j+4k)
=3i-5j+4k
Again,
W= F.X
= (5i-3j+2k).(3i-5j+4k)
= 38 joule
F= 5i-3j+2k
now, distance(x) =r2-r1
X= (5i+2j+8k)-(2i+7j+4k)
=3i-5j+4k
Again,
W= F.X
= (5i-3j+2k).(3i-5j+4k)
= 38 joule
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