1. At a water park, a 50 kg child goes down a water slide. At a pt of 3 m above
ID: 1706447 • Letter: 1
Question
1. At a water park, a 50 kg child goes down a water slide. At a pt of 3 m above the splash-down pool, the child passes a radar speedometer that determines the childs speed at 12m/s.a. Explain how the energies you calculate compare to the child's KE and PE just
before landing in the splash-down pool.
2. A ship has a mass of 40,000 kg. Its engine gives it an acceleration of 0.2m/s^2 while it travels a distance of 100 meters from rest.
A. What is the ships maximum KE after it moved 100m?
B. explain reasoning to answer for the ships KE?
Explanation / Answer
1. Well, KE + PE before = KE + PE after Now when the child is about to land in the splash-down pool, he only has kinetic energy, so it becomes KE + PE = KE .5mv^2 + mgh = .5mv^2 (.5)m(144) + m(9.8)(3) = (.5)mv^2 v = 14.24 m/s just before landing So at 3 m, the child has a certain amount of kinetic energy and potential energy, but just before he hits the pool, it is all kinetic energy and is equal to the kinetic and potential energy before at 3 m. 2. There is no change in potential energy, so Work=Kinetic Energy W=Fd=KE F=ma F=(40000)(.2) = 8000 N So Work=Kinetic Energy = 8000 x 100 = 800000 J Or you could use kinematics, since KE=.5mv^2 v^2 = v(initial)^2 + 2ax v^2 = 0 + 2(.2)(100) v= 6.325 m/s so KE = (.5)(40000)(6.325)^2 = 800000 J The Kinetic Energy is the only thing gained due to a force acting on the ship. KE=.5mv^2 so just look at either solution above for explanation.
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