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$v_f=20m/s$

Giancoli 6th Edition, Chapter 6, Problem 53

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Chapter 6, Problem 53 is solved.

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Transcript for this Giancoli solution

The initial kinetic and potential energies of this rollercoaster have to add up to the final kinetic energy and there is no final potential energy because it goes down to its reference level plus the work done by friction. And so we'll say this is one half 'mvi' squared plus 'mgh' equals one half 'mvf' squared plus the force of friction times the distance over which it operates.

We're told that the friction force is one half the weight, sorry one fifth of the weight. So that’s what that expression is, 'mg' divided by five is one fifth the weight, so that’s the friction force. And 'd' is the link to the law on the track which we're also given in the question. So we'll solve this for 'vf', we'll take this term to the left side, multiply everything by two over 'm' and then flip the sides so around so 'vf' is in the left and take the square root.

So 'vf' equals to the square root of 'vi' squared plus 'gh' over two, or I should say two 'gh'. And then minus this term here without the 'm'. so that’s two gd over five. And then we'll substitute in numbers. So the square root of, given this, that the initial speed is 1.7 meters per second so that’s squared. Plus two 9.8 times 35 meter initial height minus two times 9.8 times 45 meters is the length along the track the friction is operating. Then we get 23 meters per second for the sixth edition, for the fifth edition you have an initial height of only 30 meters instead of 35 and the fifth edition answer to this question is 20 meters per second.

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i still dont understand