Phy 201
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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I am submitting a solution to a test question for you to check. Thanks!
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 5.9 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 8 paper clips instead of one.
Since one paper clip accelerates the cart at 5.9cm/s, I am thinking we would just multiply this by 8 since the acceleration would deff have to increase. This would be 47.2cm/s.
The same system but with a single paper clip and a cart of twice the mass.
A cart of twice the mass would lead to a smaller acceleration by half. This would be 2.95cm/s
The same system but with a single paper clip with a cart of half the mass.
Half the mass would increase the acceleration. This would be 2(5.9cm/s) or 11.8cm/s
The same system but with 7 paper clips and a cart of 17 times the mass.
7 paper clips would accelerate it at 7(5.9cm/s), but the cart mass would decrease it by 17 so it would be 2.42cm/s.
What would be the acceleration of the same system but with a number of paper clips whose mass equals that of cart?
The paper clips were considered small, so in the preceding problems even though they added a bit to the mass of the cart, their proportional contribution to the mass of the system was small enough to be ignored.
However if the number of clips is large enough match the mass of the cart, we can't ignore the additional mass.
In this case a mass equal to that of the cart is suspended. If we let m stand for the mass of the cart, then the suspended mass is also m. The force of gravity, normal force and frictional force are in equilibrium on a constant-velocity ramp, so these forces do not contribute to the net force on the cart. Since gravity exerts a force F = m g on the suspended mass, the net force on the 2-mass system is m g. Its acceleration is therefore a = F_net / mass = m g / ( 2 m ) = 1/2 g.
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 7 times the cart mass?
The slope would increase when the mass is halved. This is because half the mass leads to twice the acceleration.
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OK, my only hesitation is that these paperclips are so small, so I feel like they must not have that much effect on the mass. I understand how halving the cart would lead to a huge drop in acceleration, but would we just ignore the extra mass of the paper clips in the system since they are much smaller then the cart?
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See my note, which I believe anticipated this last question. Your thinking is correct--very good.