#$&*
PHY 231
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Test 1 Logic Questions
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The below is simply questions about my logic for a few situations on a practice test I attempted tonight.
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A system consists of a cart pulled along a constant-velocity ramp by the force of gravity on a single paper clip, whose mass is much less than that of the cart, attached by a thread over a pulley with negligible friction. If the system accelerates at 4.9 cm/s2, and if F = m a describes the relationship among net force F, mass m and acceleration a, give the acceleration of each of the following:
The same system but with 6 paper clips instead of one.
The same system but with a single paper clip and a cart of twice the mass.
The same system but with a single paper clip with a cart of half the mass.
The same system but with 12 paper clips and a cart of 21 times the mass.
What would be the acceleration of the same system but with a number of paper clips whose mass equals that of cart?
How would the slope of a graph of total paper clip weight vs. acceleration for the original system (for a small number of paper clips) compare with a slope of a similar graph for a system with half the cart mass, and how would it compare for the slope of a system with 12 times the cart mass?
???I understand this problem all the way to the very end. For instance, 6x the paper clips results in 6 times the acceleration. Twice the mass results in half the acceleration. Using A= f_net/ m
The question about the slopes of the graph seem like the logic would be slightly different… they reference the slope of weight vs. acceleration. Whereas the 2nd sentence references mass of the cart(system) which would increase the acceleration to two times as much but likely have nothing to do with the clip weights. I think this means that the run (x axis) of the clips vs accel is twice as much while the clips (rise) does not change, so rise/run (the slope) is half???
@& The question is about the 'original system' to one with half the mass.
For any paper clip weight the original system would have half the acceleration of the 'half-mass' system. The acceleration of the latter system would indeed be double the acceleration of the former and the slope would be half as great.
Good thinking.*@
Next problem:
Solve using energy considerations: An automobile of mass 3.737 kg coasts up an incline which rises 1 meter for every .057 meters traveled along the incline. If the automobile has an initial velocity of 2.4 m/s, then how far along the incline will it travel before it comes to rest?
???
I understand how to find the KE which translates to work.
I know I will be using W = F * ds
Therefore, I must find force on the plane parallel to the incline. To do so I must first find the angle of the slop which is arctan(y/x)
My question comes when trying the find this which is very elementary… but wanted to verify due to the wording.. y = 1m and x = 1.75m giving the slope of .57m in the question, correct?
Once I have the force on the parallel plane the KE(work) = force on parallel plane x ds // solve for ds.
???
@& The slope is rise / run = .057 meters / (1 meter) = .057.
Technically the run is sqrt( (1 m)^2 - (.057 m)^2 ), but this is very very close to 1 m. To two significant figures is it 1 meter.
You could find the parallel component of the weight. For small slopes this component is very close to slope * weight. Then you can find the distance over which this force will dissipate the initial KE.
Alternatively you can just figure out how much higher the automobile will have to be, then figure out how far it will have to travel to get this much higher.
Gravitational force acts downward, so as the automobile rises upward gravity does negative work on it, which increases its gravitational potential energy.
The car will stop climbing when its PE has increased by an amount equal to its original KE, at which point it will have lost all its KE and be for at least an instant at rest.
*@
Next problem:
The force exerted by a rubber band at stretch x is given by the function F(x) = k x^ .71, with k = 110 N / m^ .71.
• If an object of mass .7 kg is accelerated from rest by the rubber band, which is initially stretched by .133 meters, and as a result rolls up a frictionless incline at angle 7 degrees with horizontal, then how far will the object roll along the incline (assume that none of the potential energy in the rubber band is dissipated and that the mass of the rubber band is negligible)?
???
I find the work first as the area below the slope on the Force vs stretch graph.
F(0) = 0
F(.133) = 26.26N
W = ((26.26N -0N)/2) *.133m
= 1.75 J
Was this the correct way to find force/work?
Then find the opposing force of gravity parallel to the incline.
1.73J = force parallel incline * ds
Solve for ds
Is my logic sound?
@& Almost, and you got a correct answer. But be careful about what you're thinking.
To average two forces you add them and divide by two, you don't subtract them at all.
Since one of the forces is zero this doesn't affect you answer, but be sure you are thinking clearly about how to calculate the average and what it means.*@
@& Good overall. Check my notes.*@