In the track shown in figure section AB is a quadrant of a circle of 1.0 m radius. A block is released at A and slides without friction until it reaches point B. How fast is it moving at B, the bottom of the quadrant? Share with your friends Share 8 Dinesh Pandey answered this Dear Student, Conservation of energy yields the equation,12mvB2+mghB=mghASo that,vB=2ghA-hBhA-hB=1 m ......(given)vB=2×9.8×1=4.43 m -3 View Full Answer