g Box—Continued (a) In let s(t) be the distance measured

Chapter 3, Problem 47E

(choose chapter or problem)

Sliding Box—Continued (a) In Problem 46 let s(t) be the distance measured down the inclined plane from the highest point. Use ds/dt  v(t) and the solution for each of the three cases in part (b) of Problem 46 to find the time that it takes the box to slide completely down the inclined plane. A root finding application of a CAS may be useful here.

(b) In the case in which there is friction \((\mu \neq 0)\)but no air resistance, explain why the box will not slide down the plane starting from rest from the highest point above ground when the inclination angle \(\theta\) satisfies tan \(\theta \quad \mu\).

(c) The box will slide downward on the plane when tan \(\theta \quad \mu\) if it is given an initial velocity \(v(0)=v_{0}>0\). Suppose that \(=\sqrt{3} / 4\) and \(\theta=23^{\circ}\). Verify that tan \(\theta \quad \mu\). How far will the box slide down the plane if \(v_{0}=1 \mathrm{ft} / \mathrm{s}\)?

(d) Using the values and \(=\sqrt{3} / 4 \text { and } \theta=23^{\circ}\), approximate the smallest initial velocity v0 that can be given to the box so that, starting at the highest point 50 ft above ground, it will slide completely down the inclined plane. Then find the corresponding time it takes to slide down the plane.

Text Transcription:

m neq 0

Theta

theta mu

v(0) = v_0 > 0

=\sqrt{3} / 4

u = 23^circ

v_0 = 1 ft/s

=\sqrt{3} / 4

theta=23^{\circ}