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Excerpt
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An illustration of the path independence of gravitational work.
Composition Setup
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Drop
Ramp
Track
Consider three ways to get a block from height hi down to height hf = 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 above.
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Part A
Part A
Assuming 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.
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Method 1
Method 1
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System:
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Block as plus the earth as a rigid body with infinite mass.
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In each case, there is a conservative 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.
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Model:
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.
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Approach:
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External influences from the earth (gravity) and for the block on the ramp, the ramp provides a normal force.
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Model:
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One-Dimensional Motion with Constant Acceleration and .
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External influences from the earth (gravity) and for the block on the ramp, the ramp provides a normal force.
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One-Dimensional Kinematics with Constant Acceleration.
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Part B
Part B
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Wiki Markup
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||
Assuming 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 ramp, we tilt the coordinates to place the x-axis along the ramp surface. The free body diagram for the box is:
which implies that [Newton's 2nd Law|Newton's Second Law] for the _x_ direction takes the form:
{latex}\begin{large}\[ \sum F_{x} = mg\sin\theta = ma_{x}\]\end{large}{latex}
giving:
{latex}\begin{large}\[ a_{x} = g\sin\theta\]\end{large}{latex}
With the acceleration determined, we turn to kinematics to find the final velocity. We have the givens:
{panel:title=Ramp Givens}{latex}\begin{large}\[ x_{i} = 0\]\[ x_{f} = \frac{h_{i}}{\sin\theta}\] \[v_{x,i} = 0\]\[a_{x}=g\sin\theta\] \end{large}{latex}{panel}
We then substitute into the Law of Change for one-dimensional motion:
{latex}\begin{large}\[ v_{x,f}^{2} = v_{i,f}^{2} + 2a_{x}\Delta x\]\end{large}{latex}
Substituting the givens yields:
{latex}\begin{large}\[ v_{x,f} = \sqrt{2gh_{i}}\]\end{large}{latex}
{tip}In agreement with the result from mechanical energy conservation.{tip}
{note}It is important to note the advantage of energy here. Not only does it simplify calculations in cases that can be handled by kinematics and dynamics, but it allows us to just as easily treat the case of the track, which cannot be treated with kinematics without specification of the precise path of the track.{note}
h2. Part B
In Part A, we showed that the speed of the blocks at the end of their trip is independent of the path taken from one height to another. This equality does *not* mean, however, that the trips are identical in every way. The time taken to complete the trip will depend upon the details of the path. We can show this by calculating the time taken to complete the trip for the case of the dropped box and the box sliding down the ramp using kinematics.
System: The 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 Kinematics with Constant Acceleration|1-D Motion (Constant Acceleration)].
Approach: One direct way (of several) to obtain the time is to use the Law of Change:
{latex}\begin{large}\[ v_{f} = v_{i} + at\]\end{large}{latex}
which gives:
{latex}\begin{large}\[ t_{\rm drop} = \frac{\sqrt{2gh_{i}}}{g} = \sqrt{\frac{2h_{i}}{g}}\]\[t_{\rm ramp} = \frac{\sqrt{2gh_{i}}}{g\sin\theta} = \sqrt{\frac{2h_{i}}{g\sin\theta}} \]\end{large}{latex}
{tip}As expected, the box will require a longer time to slide down the ramp than it will to drop straight to the floor.{tip}
