ACB is a smooth quater circular path of radius R. Four forces are acting at a particle placed at A. $$F1$$ is always horizontal, $$F2$$ is always vertical, $$F3$$ is always tangential to the path, $$F4$$ is always directed from particle 's position to point B. Magnitude of all forces are equal to F.

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Infinity Learn · Accepted Answer

Work done by $$F1=$$ $$F1$$ x displacement in the direction of force $$F1$$ $$=$$ $$F1$$ RWork done by $$F2$$ $$=$$ $$F2$$ x displacement in the direction of force $$F2$$ $$=$$ $$F2$$ R Work done by $$F3=$$∫$$0$$π$$/2$$ $$F3$$⋅R⋅dθ$$=F3$$×π$$R2vmax2=16$$×$$13k$$−$$39k$$−$$(13)22k=84.5k=84.5845=110(m/s)2=1000(cm/s)2$$.⇒$$vmax=1010cm/s$$

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