Correct. A child whose weight is 233 N slides down a 6.40 m playground slide that makes an angle of 37.0 with the horizontal. The coefficient of kinetic friction between slide and child is 0.0850. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.334 m/s, what is her speed at the bottom? (a) Number 101 Units J (b) Number 8.19 Units m/s eTextbook and Media Hint Solution (a) Using the force analysis we find the normal force Fy = mg cos 0 (where my = 233 N) which means fk = MKFN = HK mg cos0. Thus, AEth = fed yields AEth = fxd = ukmgd coso = (0.0850) (233) (6.40) cos37.0" = 101 J. (b) The potential energy change is AU = mg ( - d sino) = (233 N) ( - 6.40 m) sin37.0 = - 897 J. The initial kinetic energy is Ki = = my? = 2 233 N 9.8 m/ $2 (0.334 m/s)2 = 1.3J. Therefore, using W = AEme + AEt (with W = 0), the final kinetic energy is Kf = Ki - AU - AEth = 1.3 - ( - 897) - 101 = 798 J. Consequently, the final speed is of = v/2 Kf/m = 8.19 m/s