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h2. Part A

Consider three ways to get a block from height _h_~i~ down to height _h_~f~ = 0:  dropping the block straight down through the air (freefall), sliding the block down a frictionless ramp, or sliding the block along a frictionless track.  The options are illustrated below.

|!pathind1.png!|!pathind2.png!|!pathind3.png!|
||Drop||Ramp||Track||

WhatAssuming that the block starts from rest, what is the final speed of the block as it reaches the ground in each case?

Solution:  We begin by solving all three problems using [mechanical energy].  We will then re-solve the first two cases using the equations of kinematics and dynamics to illustrate the agreement between the methods.

h4. Method 1

System:  Block as [point particle] plus the earth as a rigid body with infinite mass.  

Interactions:  In each case, there is a [conservative|conservative force] gravitational interaction between the block and the earth which will provide a gravitational potential energy.  In the final two cases (the ramp and the track) the block is also subject to a [non-conservative] normal force from the surface upon which it travels.  In each case, however, the normal force does no work, since it is always perpendicular to the direction of the block's travel (the block is always moving parallel to the surface and the normal force is, by definition, perpendicular to the surface).  Thus, in all three cases, the non-conservative work is zero.

Model:  [Mechanical Energy and Non-Conservative Work].

Approach:  Since the non-conservative work is zero, the mechanical energy will be constant:

{latex}\begin{large}\[ E_{i} = E_{f}\]\end{large}{latex}

In each case, the [initial-state final-state diagram] and [energy bar graphs|energy bar graph] will be:
||        ||Initial||Final||
||\\ \\ \\Drop|!pathindinit1.png!|!pathindfin1.png!|
||\\ \\ \\Ramp|!pathindinit2.png!|!pathindfin2.png!|
||\\ \\ \\Track|!pathindinit3.png!|!pathindfin3.png!|

In each case, the constant energy relationship is:

{latex}\begin{large}\[ mgh_{i} = \frac{1}{2}mv_{f}^{2}\]\end{large}{latex}

giving:

{latex}\begin{large}\[ v_{f} = \sqrt{2gh_{i}} \]\end{large}{latex}

independent of the path taken from _h_~i~ to the zero height level.

h4. Method 2

System:  Box as [point particle].

Interactions:  External influences from the earth (gravity) and for the block on the ramp, the ramp provides a normal force.

Model: [One-Dimensional Motion with Constant Acceleration|1-D Motion (Constant Acceleration)] and [Point Particle Dynamics].

For the drop and the ramp, we can recover the same result using kinematics.

For the drop, if we set up coordinates with the positive _y_ direction pointing vertically, the acceleration will be _a_~y~ = -- _g_, so that the most straightforward approach is to use the Law of Change:

{latex}\begin{large}\[ v_{y,f}^{2} = v_{y,i}^{2} + 2a_{y}\Delta y \] \end{large}{latex}

Using the givens:

{panel:title=Drop Givens}{latex}\begin{large}\[ y_{i} = h_{i} \]\[ y_{f} = 0\] \[v_{y,i} = 0\]\[ a_{y} = - g\]\end{large}{latex}{panel}

Substituting these givens yields:

{latex}\begin{large} \[ v_{y,f} = \sqrt{2gh_{i}} \]\end{large}{latex}


For the dropramp, we tilt the coordinates to place the x-axis along the answer follows 


 ramp surface.  The free body diagram for the box is:

Giving the