Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 8 - Potential Energy and Conservation of Energy - Problems: 9d

Answer

$ 56.7m$

Work Step by Step

Total energy of the car (kinetic + potential) is: $E_{tot}=E_{kin}+E_{pot}=mgh+\dfrac {mv^{2}_{0}}{2}(1)$ We know that the total energy of the car is conserved. Therefore, at the last hill, the car will climb until its speed become zero. At this time, the kinetic energy of the car will be zero (at highest point) too so at that point, the car will only have potential energy. Now, lets write the total energy at that time: $E_{tot}=E_{kin}+E_{pot}=mgD+\dfrac {mv^{2}}{2}$ Since $v=0$, $E_{tot}=mgD\left( 2\right) $ So equating (1) and (2), we get $D=h+\dfrac {v^{2}_{0}}{2g}$ We then substitute the values into this formula to find $D$: $D=h+\dfrac {v^{2}_{0}}{2g}\approx 56.7m$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.