In our experiment you will need the mass m1 hanging vertically rather than on an
ID: 1431034 • Letter: I
Question
In our experiment you will need the mass m1 hanging vertically rather than on an incline. So, change the angle ? to 90º (make sure that you click in a different box after filling the angle value, so it is recorded.) Then click on the arrow button to release the machine: this is the type of motion that you will be studying in the lab. Based on the force vectors shown in this setup, what forces drive the motion of the mass m1 alone?
a.
The tension T1 only.
b.
The weight m1g only.
c.
Both T1 and m1g.
d.
Both m1g and m2g.
e.
Both T1 and T2.
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QUESTION 3
Release the machine and observe the motion for the following pairs of masses (m1, m2): first (80,70) and then (180,190). You may notice that the machine will accelerate differently, even though in both cases the mass difference is the same. So, how does the acceleration depend on the total mass of the system?
a.
Actually, the system does not accelerate because the forces are constant.
b.
Actually, the acceleration is the same for both mass pairs.
c.
The acceleration increases with the increasing total mass.
d.
The acceleration decreases with the increasing total mass.
e.
It depends on which is heavier, m1 or m2.
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QUESTION 4
The mass of the pulley was taken to be much smaller than the masses m1 and m2. In this case, the Atwood is said to be ideal, because the only effect of its pulley is to redirect the motion rather than contributing to the acceleration. For such an ideal pulley, the tension forces on the sides of the pulley can be assumed to be almost equal when applying the Free Body Diagram technique to find a theoretical acceleration. In turn, in lab you will measure an experimental acceleration. Which of the following is true?
a.
The theoretical acceleration will be slightly larger than experimental acceleration.
b.
The theoretical acceleration will be slightly smaller than experimental acceleration.
c.
The theoretical acceleration will be exactly the same with the experimental acceleration.
d.
The theoretical and experimental accelerations will be different (either way) only due to experimental errors.
e.
Actually, the acceleration of this system is zero because the forces are constant.
71 72 1g 12 Atwood machine Force DiagramsExplanation / Answer
2.
(c) Both T1 and m1g.
3.
m2*g - T2 = m2*a
T2 - m1*g = m1*a
m2*g - m1*g = (m1 + m2) * a
a = (m2 - m1)/(m1 + m2) * g
d. The acceleration decreases with the increasing total mass.
4.
d. The theoretical and experimental accelerations will be different (either way) only due to experimental errors.
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