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A 200-kg roller coaster car travels over the top of a 40-m high hill (Hill A) traveling at 10 m/s. It then travels down before traveling back up the next hill (Hill B) which is 20-m high. Let the gravitational potential energy be zero at the ground and assume that the work done by friction and air resistance on the car are negligible. What is the speed of the car at the top of the second hill
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conservation of total energy

P.E1+K.E1.=P.E2.+K.E2

200*g*40+.5*200*10^2=200*g*20+.5*200*v^2

=>v = sqrt(2*g*20+100)

=>v = 22.18 m/s (ans)
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