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course Phy 241
121003q001. In today's lab activity you found the work done by the rubber band used in the preceding class to energize the rotating ramp.
Give your data.
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Our rubberband chain was intially 60 cm long. We hung a plastic bag from a paperclip attached to one end of the chain and filled the bag with dominos in increments of one. Each increment we measured how long the chain was due to do force of gravity dragging the dominos down.
Chain length, Dominos.
60, 0.
63.5, 1.
67, 2.
70.5, 3.
74, 4.
79.5, 5.
84.5, 6.
91, 7.
99, 8.
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Show how you analyzed your data.
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132 g = 9 dominos.
1 dom = 14.67 g or .01467 kg.
F = m a. a = g.
F = m g.
F(dom) = .01467 kg * 9.81 m/s^2.
1 domino = .144 N.
Total work (of 8 dominos) = (.39 m) (8 dom)(.144 N/dom)(1/2) = .22 Joules.
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Give your conclusions.
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see above.
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If you have collateral observations and/or ideas for extending this investigation, give a synopsis of your observations and/or ideas.
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N/A
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`q002. Using the moment of inertia of the ramp and dominoes (found in the preceding document) and the energy in your rubber band chain, find the angular velocity of the rotating ramp, assuming that all of the potential energy stored in the rubber band chain is transferred to the ramp and that none of that energy has yet been dissipated by friction.
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Moment of inertia = .00725 kg m^2.
KE = 1/2 I omega^2.
When W nonconservative = 0, W conservative = - PE = dKE. KE 0 = 0. KE final = -PE.
Work by 8 dominoes = .22 Joules.
.22 = 1/2 (.00725)*omega^2.
omega^2 = 60.68. Omega = 7.79.
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`q003. If my truck has mass 1500 kg and descends through a total rise of magnitude 25 meters while slowing from 50 mph to 40 mph, by how much does its kinetic energy change, and how much work is done on it by the conservative gravitational force?
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40 mph = 17.8 m/s
50 mph = 22.4 ft/s.
KE point 1 = .5*(1500kg)*(17.8 m/s)^2 = 237630 N*m.
KE point 2 = .5*(1500 kg)*(22.4 m/s)^2 = 376320 N*m.
dKE = KE2 - KE1. dKE = 138890 N*m.
Work by gravity = Force *displacement. Displacement = 25 meters down, positive work.
Force = Weight* gravity = 1500kg * 9.81 m/s^2 = 14715 N.
Work by gravity = 14715 N * 25 m = 367875 N*m.
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How much work do nonconservative forces do on the truck?
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dKE = Work nonconservative + Work conservative.
dKE - W cons = W non.
W non = 138890 N*m - 367875 N*m = -228985 N*m.
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Road friction exerts a force equal to about 2% of the truck's weight. If it coasted a distance of 1000 meters along the road, what is the work done on the truck by friction?
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.02*1500 = 30 N.
-30 N (opposing motion) * 1000m = -30000 N*m.
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Friction is one of the nonconservative forces acting on the truck. What is the work done on the truck by all other nonconservative forces combined?
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Friction + Other = W non.
-30000 N*m + Other = -228985 N*m.
Other = -198985 N*m.
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What average force did those other nonconservative forces have to exert during the truck's 1000 meter displacement?
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Work = fAve * Displacement.
fAve = 198.985 N.
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My 2000 kg station wagon coasts along the same path and its speed decreases from 50 mph to 47 mph. Friction again exerts a force equal to 2% of the vehicle's weight. What average force was exerted by nonconservative forces other than friction?
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`q004. University Physics.
You did an experiment with a balanced vertical strap, to which two small magnets (mass 3 grams each) were added. The goal was to determine how much energy was transferred from the rubber band chain to the strap.
Give your data.
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We spotted angle theta that the top half of the strap made with the level table.
We timed the oscillations.
oscillations, time, angle theta in degrees.
10, 31, 35.
20, 55, 15.
30, 1:18, 5.
40, 1:40, 0.
50, 2:01, -5.
1 arm length = 15 cm.
We could then therefore use the height 15 cm + 15sin(theta) to determine displacement vertically of the magnets. From this we can calculate change in potential energy between two points.
Also, knowing Theta and Time intervals, we can calculate omega.
From omega and Moment of inertia (.0006225 kg*m^2 from previous data) we can use KE = 1/2 I omega^2 to find kinetic energy. We can find change in kinetic energy between two points.
Total Work = Work noncons + Work cons.
- Work cons = PE.
Work noncons - PE = dKE. Thus we can calculate Work noncons.
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Good.
It appears that the period of oscillation decreases fairly steadily from 3.1 seconds to about 2.4 seconds. Can you confirm this, and do you see any interesting trends?
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Show how you analyzed your data.
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see above
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Give your conclusions.
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see above
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If you have collateral observations and/or ideas for extending this investigation, give a synopsis of your observations and/or ideas.
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`q005. University Physics.
You were asked to think about how you could determine the energy dissipated per centimeter by rolling friction on the axel.
You were advised to think of everything that could be measured for that system.
Please share your thoughts, and if possible, address the question of how some of the things that could be measured might pertain to the original goal of determining energy dissipated by rolling friction.
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From above, the Nonconservative forces are friction (rolling) and wind resistance. Wind resistance is negligible, so Rolling Friction = Noncons. Therefore Work done by Rolling Friction = Work Noncons.
Displacement = radius* theta.
Knowing theta, and measuring the radius of the axle, we could determine linear displacement.
Here is where my brain stops working...
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That's pretty good.
You could construct a graph of angular amplitude vs. clock time.
You could construct a graph of period vs. angular amplitude.
From these graphs you could construct a graph of average angular speed vs. time.
Integrating this graph you could determine the angular distance traveled by the system.
From the radius of the axle you could then determine the rolling distance.
From that and the energy loss you could find the average force of rolling friction.
Spend some reasonable amount of time thinking about this and see if you can come up with a decent set of graphs.
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There are other alternatives, along the same line of thinking, some perhaps a little simpler. You're welcome to explore those alternatives if you prefer.
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Pretty good.
Give yourself 30-40 additional minutes on my questions and see what you can come up with.
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