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A constant force `F=(hati+3hatj+4hatk)N` acts on a particle and displace it from (-1m,2m,1m)to(2m,-3m,1m). |
Answer» Given, `F=hat i+3hatj+4hatk` and Initail position of the particle is given by `s_(1)=-hat i+2hatj+hat k` Final position, `s_(2)= 2hati-3hatj+hatk` Displacement of the particle , `s=s_(2)-s_(1)=(2hati-3hatj+hat k)-(hat i+2hatj+hat k)` ltbrlt `=3hati-5hatj+0hatk=3hatj-5hatj` Work done by the force F `W=F.s=(hat i+3hatj+4hatj).(3hati-5hatj)=(3-15)=-12 J` |
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