course Phy121
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14:45:33
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RESPONSE -->
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14:46:15 Why is the slope of the graph obtained in this experiment equal, or very nearly equal to the acceleration of gravity?
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RESPONSE --> Because the slope of the graph is the acceleration...and gravity is the acting force upon the cart
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14:46:23 ** The gravitational force component down the incline, for small slopes, is proportional to the slope in such a way that this component is equal to the product of slope and object weight. The acceleration due to this force would in the absence of friction be equal to the product of the slope and the acceleration of gravity (slope * object weight / object mass = slope * acceleration of gravity, since weight = mass * accel of gravity). The opposing frictional force is very nearly constant, which shifts the 'ideal' frictionless graph horizontally by a uniform amount. This doesn't affect the slope. So the slope of the graph is the same as if we graphed slope * accel of gravity vs. slope; that slope is accel of gravity. **
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RESPONSE --> ok
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14:49:29 Why does the linearity of our force vs. slope data and of the slope vs. acceleration data imply that acceleration vs. force is linear?
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RESPONSE --> Because gravity is acting upon the cart...Gravity produces the force depending upond the slop, and also acceleration of gravity depending upon the slope. Thus the acceleration and the force are both gravity dependent acting upon the object
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14:49:36 ** Quantities that are linearly related to the same quantity are linearly related to one another. Both force and accel are linearly related to slope, so force must be linearly related to accel. The rigorous explanation is mathematical: If y = m x + b and x = n z + c then y = m ( n z + c ) + b = mn * z + ( m c + b ), which is a linear function of z. If y is a linear function of x and x is a linear function of z, then y is a linear function of z. So if force is a linear function of slope and slope is a linear function of acceleration then force is a linear function of acceleration. **
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RESPONSE --> ok
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14:53:18 How do we then determine that acceleration is proportional to force?
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RESPONSE --> The relationship of gravity acting upon an object depends upon the force of which gravity acts upon. Using the equation F=ma can be applied to this solution
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14:53:29 ** Accel proportional to force means accel linearly proportional to force. A linear proportionality means that accel = k * force, where k is a constant number for the object being accelerated. The linear proportionality is represented by a linear graph thru the origin. If accel is linearly related to force and if the graph of accel vs. F passes thru the origin, within margin of error, then we have a linear proportionality. **
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RESPONSE --> ok
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14:56:28 QUESTION FROM STUDENT--Please define the differnece between Fnet and Force. See if you can answer this question.
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RESPONSE --> Fnet is the vector sum of all of the forces acting upon an object. Where Force is a general term used for the force applied to an object
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14:56:37 ** Net force is the sum of all forces acting on an object. If you're pushing your car you are exerting a force, friction is opposing you, and the let force is the sum of the two (noting that one is positive, the other negative so you end up with net force less than the force you are exerting). Your heart rate responds to the force you are exerting and the speed with which the car is moving; the accel of the car depends on the net force. **
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RESPONSE --> ok
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14:57:38 In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that the change in v^2 is proportional to Fnet `ds?
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RESPONSE --> By rearrainging several of the formulas we have for force and motion...we can show this relationship quite distinctly
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14:57:50 ** It's very important in physics to be able to think in terms of proportionality. To say that y is proportional to x is to say that for some k, y = k x. That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that for some k, a * `ds = k * ( change in v^2)--i.e., that a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for k = 1/2. Now since Fnet = m a we see that Fnet is proportional to a for a given mass m, and it follows that Fnet `ds = k * change in v^2, for the appropriate k (specifically for k = mass / 2. **
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RESPONSE --> ok
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14:58:28 How do our experimental results confirm or refute this hypothesis?
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RESPONSE --> The experimental data is does confirm these results within reason.
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14:58:38 ** We didn't actually do this part of the experiment, but on a ramp with fixed slope a `ds is simply proportional to `ds. When we measured `dt for different distances from rest down the same ramp, we were then able to determine the average and final velocities. The change in v^2 for each timing would be from 0 to vf^2. The change would therefore be just vf^2. If a `ds is proportional to the change in vf^2 then a graph of vf^2 vs. a `ds should be linear. Since a is constant we don't even need to determine it--a graph of vf^2 vs. `ds would be linear. This would confirm the hypothesis. **
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RESPONSE --> ok
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15:09:24 video experiment 6: What horizontal velocities did you get on experiment 6?
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RESPONSE --> See Report
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15:10:15 ** Results generally confirm horiz velocities between 80 cm/s and 85 cm/s, which within experimental error confirms that horiz velocity is constant. To confirm this the graph would of v vs. vertical fall will not only have to be linear, but would have to be consistent with the hypothesis that the line is horizontal. Within reasonable error estimates, this is the case for the graphs usually obtained. **
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RESPONSE --> Seems consistent with my data which was obtained at 84 cm/s
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