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 - Page 207: 65


The particle stops at $20$ $cm$ from the left end of the flat part.

Work Step by Step

Its Potential energy at point $A$ is $m\times g\times h=mgh$ Because the surface is frictionless, the energy is conserved and is converted to kinetic energy at bottom, just before the flat part. So, $Potential$ $Energy =Kinetic$ $Energy$ $mgh=\frac{1}{2}mv^{2}$ $ v^{2}=2gh$ Substituting the known values in this expression gives: $v^{2} =2\times g\times 20$ $v^{2}= 40g $ Now, acceleration due to friction is $\frac{Frictional force} {mass}$ $\frac {0.2 mg}{m}=-0.2g$ $m/s^{2}$ Negative sign denotes opposite direction. We know, $v^{2}=u^{2}+2ad$, Plugging the known values and solving gives: $0^{2}=40g+2(-0.2g)d$ $d=100$ This means it will travel a total distance of $100cm$ throughout its motion along the flat part. Thus, it will go to the other end and comeback covering $40cm\times 2=80cm$ and then cover the remaining $20cm$ from the the left edge. Therefore it stops at $20$ $cm$ from the left edge.
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