Question

5. A 65.0-kg woman steps off a 10.0-m diving platform and drops straight down into the water. After entering the water she continues falling through the water until she reaches a depth of 4.00 m. a. Calculate the work done on the woman by gravity and by the water. Ignore air resistance. b. What is the average net force exerted on her over this 4.00-m displacement in the water? c. How does the average net force exerted on the woman while she is moving downward in the water compare to the force exerted on her by the water during this dis placement?

Answer 1

solution (ay Work done by gravity WG = Mgh OR WG = 6510 x 9,8 x (10.0+400) = 6570 x 9,8 x 1400 = 8918 J WG =8918 7 J Work done by water = WWater - 89185 Whater = 1-8918 (6) Fang x 4.00 = 8918. > Faug = | 2229.5 IN Bio = -24,5 m/s2 (c) acceleration al woman in the Water = o-vi2 - 2xuico but Vi = 129.8 1010 = (2x9.8x10.0 = 1400 mis so a = -14.022 But F-ng = mai =) F= M (9+a)= 6510X(9.8+245) "CO = 2229,5 so both forces that s during the motion in the water downward and average force for the displacement are equal Both are equal