Physics, asked by jimmy990, 1 year ago

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

Answers

Answered by Adithiya2002
25
Integration of force with respect to displacement gives Work done.

By following the above statement,
we get,

a)1/6J

Regards,
Adithiya
Answered by mehtaprag03
40

Answer:

Explanation:

The answer is given below !!

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