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course Phy 231
10/23/2011 @ 10:34p.m.
New Physics_VideosAs discussed in class you should master Introductory Problem Set 5 before next Monday's class.
Check the document at Notes and comments on assignment due 111012 for notes, comments and hints on the problems you submitted for 111012. If you haven't submitted those problems, do so before you look.
After looking, use the comments and hints, along with the notes as posted in your access page, to revise your work and submit your revisions.
Your work on nswers to the following questions will be due Wednesday, 10/19/11. The first four questions look like those assigned on 111003 but they are different.
Some questions, as usual, are more challenging than others. Concentrate first on the questions you can answer most easily. Then go back and try the ones you find a little more challenging.
`q001. For each of the given objects on the various inclines find the parallel and perpendicular components of the object's weight as a percent of its weight, using the sine and cosine functions. Compare with the estimates you made previously.
A car weighing 20 000 Newtons on an incline making angle 12 degrees with horizontal.
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Signs assume down the ramp being positive for parallel and the down direction positive for perpendicular.
Fpar = 4,158N (21%, estimate was 30%)
Fper = 19,563N (98%, estimate was 95%)
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A hotwheels car of weight 80 000 dynes on an incline whose angle with horizontal is 20 degrees.
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Fpar = 27,362 dynes (34%, estimate was 50%)
Fper = 75,175 dynes (94%, estimate was 90%)
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A block of weight 30 pounds on a 37 degree incline.
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Fpar = 18lb (30%, estimate was 30%)
Fper = 24lb (80%, estimate was 70%)
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`q002. Your answers to this question will be based on the results you got using the sines and cosines, and will differ from those you obtained previously using estimates. If the car in the first question experiences a frictional force which is 2% of the perpendicular component of its weight, then what is the magnitude of the frictional force?
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Ffrict = -391N
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If the car is coasting downhill what is the sum of the parallel component of its weight and the frictional force?
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Fpar = 3,767N
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If the car is coasting uphill what is the sum of the parallel component of its weight and the frictional force?
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Fpar = -4,549N
I just realized that on my estimates assignment, I used friction and perpendicular component to find fparallel instead of friction and parallel components…..
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`q003. Again your results will be based on the calculations you did in the first question, and should be compared to your previous estaimtes.
If the hotwheels car in the first question is attached by a light thread to a washer weighing 20 000 dynes and suspended over a pulley at the lower end of the ramp, then if friction is ignored what is the net force acting in the direction down the ramp?
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47,362 dynes (80% of my estimate)
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Answer the same question assuming that the washer is suspended from a pulley at the top of the ramp.
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7,362 dynes (35% of my estimate)
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`q004. Once more base your answers on the results obtained for the first question, and compare your results with your previous estimate-based results.
How much frictional force would it take to hold the block in the first question stationary on the incline?
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-18lb (same as my estimate)
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What is this force as a percent of the weight of the block?
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60% (same as my estimate)
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`q005. Sound in air travels at about 340 m/s. If you drop a rock down a well which is 40 meters deep, how much time will elapse between the drop and hearing the sound of the splash?
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Using ‘ds = v0’dt + ½a’dt^2, I found ‘dt for the rock to hit to be 2.86 seconds, and then through ‘ds/vAve I found time for sound to go up from the bottom to be .12 second. Adding these times gives a total ‘dt of 2.98 seconds…all of this assumes a for sound is ignored since it probably reaches its max velocity almost instantly.
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`q006. A rock of mass 2 kilograms is tossed upward at 6 meters per second, being released at a height of 4 meters above the ground. Assume that air resistance has no significant effect on its motion.
How long after being released will it reach the ground?
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Using vf^2 = v0^2 + 2a’ds, ‘ds = 1.84m.
Since the object stops at the top and has constant a, vAve = 3m/s, so ‘dt to the top = .61seconds.
From here, the rock will fall 5.84m and using ‘ds = 1/2a’dt^2, we can find the ‘dt from the top to the ground to be 1.09 seconds, which means a total ‘dt of 1.7 seconds.
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@& You don't have to separate the upward motion from the downward.
v0 = 6 m/s, `ds = -4 m and a = -9.8 m/s^2 directly yield the time, using the third equation of unif accel motion.*@
What will be its kinetic energy at the instant it reaches the ground?
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The vf = (9.8m/s^2 ^ 1.09 seconds) which is 10.68m/s.
KE when it hit was 114J,
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What was its kinetic energy when just after its release?
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36J
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How much work did gravity do on the ball between release and striking the ground?
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Work during its ascent was -36J, during its fall was 114J, for a total work of 78J.
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@& `ds = 4 m down, grav force is 19.6 N down, so work by gravity is 78.2 J.
No need to consider ascent and descent separately. The force is conservative, therefore path-independent. Only initial and final positions matter.*@
How much work did gravity do on the ball between release and reaching its maximum altitude?
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-36J
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@& Now of course it's necessary to calculate work going up.*@
How much work did gravity do on the ball as it fell from its maximum altitude to the ground?
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114J
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`q007. A hotwheels car of mass 50 grams rolls down a ramp inclined at 30 degrees from horizontal.
How much force does gravity exert on it?
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490g*cm/sec^2
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What is the component of the gravitational force parallel to the incline?
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245g*cm/sec^2
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If friction is negligible, what therefore will be the acceleration of the car?
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A = fpar / m = 4.9m/s^2
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`q008. A rubber band breaks when its tension reaches 20 Newtons. The rubber band is used to accelerate a 1.2 kg mass upward, with gradually increasing acceleration. When the acceleration reaches a certain value the rubber band breaks. What is that acceleration?
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F = m * a
20N = 1.2kg * a
A = 16.67m/s^2
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@& To accelerate the mass at 16.67 m/s^2 the net force must be 20 N.
The net force is the sum of the tension and gravity forces. Tension is up and gravity is down, so tension is significantly greater than net force.
Meaning tension at that acceleration is significantly greater than 20 N.*@
`q009. Two dominoes are stacked, one on top of the other. The maximum frictional force exerted between the dominoes is 10% of the weight of the top domino. The bottom domino is attached to a rubber band chain, which is stretched to some length before the dominoes are released. This is repeated, slightly increasing the length of the rubber band chain with each repetition, until the top domino slides off.
What acceleration was necessary to cause this to occur?
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.1 * 15g = 1.5
1.5 = 15g * a
A = .03m/s^2 (I not only do not feel comfortable about my approach, but my units seem inaccurate as well)
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@& You should be uncomfortable with the fact that you have no forces in your analysis. Units aren't rigorously used, which leads to errors.
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Will the top domino begin to slide shortly after the moment of release, or does this occur a bit later?
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It will begin to slide slightly after due to the bottom domino decelerating faster than the top domino.
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`q010. A spring exerts an average force of 3 Newtons on a 100-gram toy car, initially at rest, as the car moves 10 centimeters. Other forces in the direction of the car's motion are not significant, nor is the mass of the spring.
How fast will the car be moving at that point?
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3N = .1kg * a
A = 30m/s^2
Using ‘ds = 1/2a’dt^2, I found ‘dt to be .08seconds
This means vf = 2.4m/s
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@& Better to figure work done by the force, and apply work-KE theorem.
100 gram mass with KE of .3 Joules, etc..
The result is about 2.4 m/s, but no need to analyze the motion.*@
Assuming that the spring's force is conservative, by how much did its potential energy change during this interval?
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‘dKE = .288J, which means the PE could change by .288J.
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@& `dKE = .3 Joules, from work-KE theorem.
You're off by the rounding of sqrt(6) to 2.4. More like 2.45, which would give you KE closer to .3 J.*@
`q011. A 100-gram steel ball observed to be moving at 80 cm/s collides with a marble of mass 30 kg, which is initially at rest. Immediately after the collision the ball is observed to be moving at 60 cm/s and the marble at 66 cm/s.
What is the total momentum of the two objects before collision?
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Pball = .08kg*m/sec, pmarble = 0kg*m/s
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What is their total momentum after collision?
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Pball = .06kg*m/sec, pmarble = 19.8kg*m/sec
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What was the change in the momentum of the steel ball?
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-.02kg*m/sec
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What was the change in the momentum of the marble?
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+19.8kg*m/sec
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How much total kinetic energy was present before the collision?
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.032J
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How much total kinetic energy was present after the collision?
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6.518J
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Estimate how long the steel ball and the marble would have been in contact.
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.1sec
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Based on your estimate what average force was exerted by the ball on the marble, and what average force by the marble on the ball?
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Fball = .2N, Fmarble = 198N (on the ball and on the marble)
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`q012. An object moving at velocity v along the arc of a circle of radius r is being accelerated toward the center of the circle, with acceleration a_centripetal = v^2 / r. Note that 'centripetal' means 'toward the center'.
A domino of mass 15 grams is balanced each end of a metal strap of length 40 cm, which is rotating about its center at a rate that gives the domino a speed of 10 cm / second.
What is the centripetal acceleration of the domino?
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Cent_A = 100cm^2/s^2 / 20m = 5cm/sec^2
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What is the centripetal force on the domino?
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Cent_F = 15g * 5cm/sec = 75g*cm/sec
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Where does this centripetal force come from?
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The force comes from the string pulling the object inward (accelerating it), which causes an object to continue in a circular path rather than going straight.
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How fast would the domino have to be moving in order for its centripetal acceleration to be 1.5 m/s^2?
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.55m/sec
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How fast would the domino have to be moving in order for the centripetal force to be 10% of its weight?
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1.4cm/sec
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*#&!
&#Your work looks good. See my notes. Let me know if you have any questions. &#