A force acting on a particle varies with the displacement x as F=ax-bx^2.Where a=1 N/m and b=1 N/m^2.The work done by this force for the first one meter (F is in newtons,x is in meters) is:
(a) 1/6 J (b) 2/6 J (c) 3/6 J (d) None of these
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Answered by
25
Integration of force with respect to displacement gives Work done.
By following the above statement,
we get,
a)1/6J
Regards,
Adithiya
By following the above statement,
we get,
a)1/6J
Regards,
Adithiya
Answered by
40
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