A 1000kg car is traveling at 20 m/s at the top of hill that is 50 m high. The ca

Question

A 1000kg car is traveling at 20 m/s at the top of hill that is 50 m high. The car then coasts down the hill and up the next hill with a height of 50m high. Friction and air resistance combined do 1.25 Times 105 Joules of Work? Will the car reach the top of the second hill If the car will reach the top of the second hill what will the speed of the car be at the top of the second hill ? If the car does not reach the top of the second hill, what height will the car reach before it stops? If the car does reach the top of the second hill, what is the maximum height of hill that the car could reach?

Explanation / Answer

KE1 = (0.5*1000*20*20) = 2e5J

PE1 = m*g*h = (1000*9.8*50) = 4.9e+5

initial energy = E1 = KEi + PEi = 6.9e+5 J

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if the reaches the top of the second hill PE2 = m*g*h = 4.9e+5 J

after the car coast the down the hill and up the next hill

energy E2 = E1 - Wf = 690000 - 1.25e5 = 5.65e5 J

E2 > PE2

yes the car will reach the top of the hill

let at the top the speed is v2

KE2 = 0.5*m*v2^2

E2 = PE2 + KE2

5.65e5 = 4.9e+5 + 0.5*1000*v2^2

v2 = 12.25 m/s <------answer

++++++++++++

part b)

E2 = PEmax

5.65e5 = 1000*9.8*h

h = 57.65 m <-----answer

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