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: 65

Answer

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.
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.