060218_notes11questions

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20:24:06 How do the experiment with acceleration vs. number of washers and experiment with acceleration vs. ramp slope convince us that for a given cart there is a net force down the ramp which is proportional to the slope in excess of the constant-velocity slope (the slope excess)?

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RESPONSE --> makes a straight line on a graph

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20:24:19 ** A graph of acceleration vs. the number of washers in excess of the number required to maintain a constant velocity is consistent with a straight-line graph through the origin. **

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RESPONSE --> yes it does goes through the origini

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20:27:52 If for small slopes we find that the net force exerted on a cart is proportional to the slope excess, then why does Newton's Second Law ensure us that for small slopes the acceleration of the cart should in fact be proportional to the slope excess?

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RESPONSE --> because the law says that for a fixed mass the acceleration is proportional to the force, so if for small slopes force is proportional to slope and acceleration is proportional to slope

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20:27:59 ** Newton's Second Law tells us that for a fixed mass the acceleration is proportional to the force, so if for small slopes force is proportional to slope excess, then acceleration must also be proportional to slope excess. **

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RESPONSE --> ok

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20:29:23 How is the slope of the net force vs. acceleration graph for an object reveal the mass of the object? Why should a greater mass result in a greater slope?

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RESPONSE --> Force equals mass time acceleration. F=m*a Which the graph would then be F vs. a and you can get the slope m

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20:29:32 ** Since F = m a, when we graph F vs. a the slope is m. **

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RESPONSE --> right

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20:32:56 How do we obtain our definitions of work and kinetic energy in terms of the equation vf^2 = v0^2 - 2 a `ds and Newton's Second Law?

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RESPONSE --> a = Fnet/m you put this into the equation vf^2 = v0^2 + 2*a*'ds then set up the equation for Fnet * 'ds....

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20:33:46 ** Substituting a = Fnet / m into vf^2 = v0^2 + 2 a `ds we get vf^2 = v0^2 + 2 Fnet / m * `ds, which we easily rearrange to get Fnet `ds = 1/2 m vf^2 - 1/2 m v0^2. Fnet `ds is defined as work, 1/2 m v^2 is defined as KE, and our equation tells us that the work done by the net force is equal to the change in KE. **

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RESPONSE --> yes, Fnet *'ds is work. KE is 0.5*m*v^2

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Also good. Let me know if you have questions.