A force F = 20 + 10y acts on a particle in y-direction, where F is in newton and y is in metre.  Work done by this force to move the particle from y = 0 to y = 1 m is [NEET 2019]

Work done by a force F, which is variable in nature in moving a particle from y1 to y2 is given by  W=∫y1y2F·dy  … (i) Given, force,  F=20+10y,y1=0 and y2=1 mSubstituting the given values in Eq. (i), we getW=∫01(20+10y)dy=20y+10y2201=20(1-0)+5(1-0)2=25∴   Work done by the force will be 25 J.