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course Phy201
Here are some questions related to the 101025 class. Notes will be completed soon.
`q001. If the acceleration of an Atwood system with total mass 80 grams is 50 cm/s^2, then:
How much mass is on each side? Note that this can be reasoned out easily without an complicated analysis, using the same type of reasoning that led us to conclude that a system with 31 g on one side and 30 g on the other accelerates at 1/61 the acceleration of gravity.
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Mass on each side wouldn’t it be like you said 31g and 30g
The idea applies to this question; however the details in this problem are different.
In this case the acceleration is 50 cm/s^2, which is 1/20 the acceleration of gravity.
How would the 80 grams have to be 'split up' in order to result in acceleration 50 cm/s^2?
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By analyzing the forces on the mass on the 'lighter' side, what is the tension in the string?
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f=ma
F=30g(50cm/s^2)
f=1500J
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By analyzing the forces on the mass on the 'heavier' side, what is the tension in the string?
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F=ma
F=31g (50cm/s^2)
f=1550J
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`q002. A large sweet potato has mass 1188.6 grams. Sweet potatoes sink in water. Suppose that when this sweet potato is suspended as the mass one one side of an Atwood machine, but immersed in water, a mass of 100 grams on the other side is required to balance it.
Sketch the forces on that system and describe your sketch.
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Well mg is straight down
The tension would also be straight down
The tension in the string is pulling up on the sweet potato, on one side, and on the 100 gram mass on the other side. Both tension forces are upward.
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Would the sweet potato, if reshaped without changing its volume, fit into that 1-liter container?
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No it would be over because of the buoyant forces is smaller than the weight force by around 10% so the weight vector is more and would be hanging out.
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What if the required balancing mass was 200 grams?
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Well the forces would all be doubled.
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`q003. The balloon rose to the ceiling in about 1.5 seconds, 2 seconds, 2.5 seconds and 5 seconds when 1, 2, 3 and 4 paperclips, respectively, were attached. It fell to the floor in about 6 seconds when 5 paperclips were attached. Assume the displacement to have been the same in each case. The buoyant force results from the fact that air pressure decreases as altitude changes, which results in more force from the air pressure on the bottom of the balloon than on the top. The pressure in the room changes at a very nearly constant rate with respect to altitude, so the buoyant force can be assumed to remain constant throughout the room.
Assuming uniform acceleration in each case, is a graph of acceleration vs. number of clips linear?
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Yes it is linear.
you need to do some calculations to support your conclusion, and the calculations should be related to the acceleration
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Do your results indicate the presence of a force other than the gravitational and buoyant forces acting on the balloon?
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Well other than the fan that was blowing the room no I did not have any other forces.
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`q004. When a certain object coasts up an incline its acceleration has magnitude 100 cm/s^2 and is directed down the incline. When it coasts down the incline its acceleration is 50 cm/s^2 and is directed up the incline. Only gravitational, normal and frictional forces are present.
Sketch a figure depicting the forces on the object as it coasts up, and as it coasts down the incline. Describe your sketch.
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Well when the object is going down the incline the mg is straight down and friction is positive direction.
Well when the object is going up the incline the mg is still straight down and the friciton is in the negative direction.
Goth answers are correct, assuming that the positive direction is up the ramp.
If the weight vector m g, which is indeed straight down, is broken into its components parallel and perpendicular to the ramp, you get an even better picture.
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Are the magnitudes of the force vector depicted in your sketch consistent with the given accelerations? If not, make another sketch and adjust the vectors as necessary. Then describe why you think your sketch is a reasonable representation of the system.
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Yes I think it is reasonable because of the acceleration are different so the frictional forces would be different.
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From the given information you can determine the coefficient of friction. You may assume that the normal force varies little in magnitude from the weight of the object. What is the coefficient of friction?
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Coefficient of friction .03
That doesn't follow. Check the Class Notes for 111003.
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Having determined the coefficient of friction, you can also determine the slope of the incline. For simplicity you can assume that since the slope is small, the magnitude of the weight component parallel to the incline is equal to the slope multiplied by the weight of the object. What do you get?
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.05
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`q005. If parts of the preceding problem gave you trouble, consider an object on an incline with slope .05 and coefficient of friction .03. You can again assume that since the slope is small, the magnitude of the weight component parallel to the incline is equal to the slope multiplied by the weight of the object, and also that the normal force does not differ significantly in magnitude from the object's weight. Let m stand for the mass of the object, g for the acceleration of gravity.
In terms of m and g:
What is the weight of the object?
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w=mg
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What is the magnitude of its weight parallel to the incline?
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-.05mg
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What is the magnitude of the normal force?
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.03mg
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What is the magnitude of the frictional force?
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.015g
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What is the magnitude of the net force when the object coasts up the incline?
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Fnet=-.05mg-.03mg
Fnet=-.08mg
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What is the magnitude of the net force when the object coasts down the incline?
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Fnet=-.05mg+.03mg
Fnet=-.02mg
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What therefore are the object's accelerations up, and down, the incline?
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.045
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What are those accelerations in cm/s^2?
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.045cm/s^2
very good work on this question
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&#Good responses. See my notes and let me know if you have questions. &#