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 $$2019$$]

Question

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 $$2019$$]

Infinity Learn · Accepted Answer

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.

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]

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