A child weighing 50kg is pushed down a 2m long slide (u=0.1), angled at 45 degrees from the horizontal, at 5m/s. At what speed does the child reach the bottom of the slide?

Change in GPE=Change in KE + ResistanceResistance = ForcexDistnaceForce=Friction= R u = masscos45g2 * 0.1Change in KE = 0.5m*(v²-25)Change in GPE = mg2Sin 45Can cancel down by m (both sides in all parts)9.81.414=0.5(v^2-25)+0.14149.82(9.81.414 -0.14149.8)=(v²-25)v²= 49.94v=7.06 m/s