Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
Nov 2013 p43 q1
4025
A particle moves up a line of greatest slope of a rough plane inclined at an angle \(\alpha\) to the horizontal, where \(\sin \alpha = 0.28\). The coefficient of friction between the particle and the plane is \(\frac{1}{3}\).
Show that the acceleration of the particle is \(-6 \text{ m s}^{-2}\).
Given that the particle’s initial speed is \(5.4 \text{ m s}^{-1}\), find the distance that the particle travels up the plane.
Solution
(i) Using Newton's second law and the equation for friction \(F = \mu R\), we have:
\(-\left(\frac{1}{3}\right)(W \cos \alpha) - W \sin \alpha = \left(\frac{W}{g}\right)a\)
Substituting \(\sin \alpha = 0.28\) and \(\cos \alpha = \sqrt{1 - \sin^2 \alpha} \approx 0.96\), we get:
\(-(0.32 - 0.28)g = a\)
\(a = -6\).
(ii) Using the equation \(0 = u^2 + 2as\) where \(u = 5.4 \text{ m s}^{-1}\) and \(a = -6 \text{ m s}^{-2}\):