A man, who weighs 190 lb, slides down from the top of the water slide. His initi

Question

A man, who weighs 190 lb, slides down from the top of the water slide. His initial speed is zero, there is no friction between the slide and the rider, the height of the slide (height difference between A and B) is hi, and the distance between the bottom of the slide and the water surface level is hi, and the slope of the water slide at the top and the bottom is zero. Assume that h1 = 12 ft, h2 = 8 ft, and L=26 ft. Determine the speed of the rider al the moment he becomes airborne (at location B).

Explanation / Answer

1) Using conservation of energy,

Loss in Potnetial Energy = Gain in Kinetic Energy

m*g*h1 = 0.5*m*v^2

190 * 12 = 0.5 * (190/32.2) * v^2

v = 27.8 ft/s

2) Work done by gravity = dot (F*dr)

Force and displacement are in same direction, so work done would be positive

W = m*g*(h1+h2) = 190*20 = 3800 ft.lb

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