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$v=60.2m/s$

Giancoli 6th Edition, Chapter 6, Problem 34

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Chapter 6, Problem 34 is solved.

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Transcript for this Giancoli solution

We know that the size of the changing kinetic energy is going to be the same as the amount of change in the potential energy. And the angle of the slope, as a matter, the only thing that matters for this gravitational potential, is the actual change in height. Since gravity is straight up and down, and so is height. So we have one half 'mv' squared equals 'mgh'. In solving for 'v', and the 'm's will cancel, which is nice. And we'll multiply both sides by 2, and take the square root of both sides. So 'v' is the square root of 2'gh'. And this becomes, square root of 2, times 9.8 time 185 Meters, which equals # meters per second. In the 5th edition, this 185 meters, is actually 125 meters. So for the 5th edition the final speed is instead # meters per second.

Comments

I'm curious, why didn't you use the actual height of the slope rather than the length of the hypotenuse? Similar to what you did with question 49