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course phys201
`q001. A domino of mass .02 kg is moving at 140 cm / second. What is its kinetic energy?½.02kg(1.4m/s)^2= .019J
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If the kinetic energy of a domino of mass .02 kg is .5 Joules, how fast is it moving?
.5=1/2(.02kg)x^2
X=7.07m/s
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An object of unknown mass is given a kinetic energy of 8 Joules, which causes it to move at 12 meters / second. What is its mass?
8=½(x)12m/s^2
=.03kg
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The equation would be
8 Joules =½(x)(12m/s)^2.
Note the added sign of grouping, which distinguishes the square of the 12 m/s velocity from an acceleration of 12 m/s^2.
Expressing the Joules in fundamental units you get
8 kg m^2 / s^2 =½(x)(12m/s)^2.
The solution would be x = 0.11 kg, not .03 kg.
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All the above questions would be answered using a single definition. What is that definition?
KE=1/2mv^2
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`q002. A domino of mass .02 kg is positioned on a rotating strap, 20 cm from the axis of rotation. How far does the domino move as the strap completes a full rotation, and how far does it move as the strap rotates through 220 degrees?
Struggled with this question, unsure of the best method to go about to solve this problem
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The domino moves around the circumference of a circle.
So use basic facts about the circle.
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How fast is the domino moving if the strap is rotating at 220 degrees / second, and what is its kinetic energy at that speed?
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Through how many degrees per second would the strap have to be rotating in order for the domino to have 0.1 Joules of kinetic energy?
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These questions would be answered with the same definition used for the first question, in addition to what common knowledge from high school mathematics?
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`q003. A horse of mass 500 kg starts from rest and descends a steep bank, losing 50 000 Joules of gravitational potential energy. How fast would the horse be moving if all the lost potential energy was converted to kinetic energy?
50,000=1/2(500)V^2
V=+- 14.1
dKE= 50,000
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If at the bottom of the bank the horse in the preceding is moving at 10 meters / second, how much work was done on it by nonconservative forces?
½(500kg)(10m/s)^2
dKE=25,000
dW= 25,000 J
500kg(9.8m/s^2
=20,100J
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500kg(9.8m/s^2) isn't 20,100 J.
Is your `dW the work done by the net force, the work done by conservative forces or the work done by nonconservative forces? It isn't clear from just the answer how you reasoned this out.
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What might be the nature of these nonconservative forces?
Friction, air resistence
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The first two questions were answered based on what definitions and what theorems?
KE=1/2mv^2, dW=dKE, work energy theorem
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Good, but the work-energy theorem says that
`dW_net = `dKE,
not just `dW = `dKE. It's essential to distinguish the work done by net, conservative and nonconservative forces.
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`q003. What is the kinetic energy of a rotating system whose moment of inertia is .012 kg m^2 when it is rotating at 4 radians / second?
KE=1/2 .012kgm^2(4r/s)^2
KE=.096J
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What is the angular velocity of this rotating system if its kinetic energy is 0.6 Joules?
.6=1/2(.012kgm^2)w^2
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omega, not w.
Spell it out. If for no other reason, w has too many other connotations and use of w for omega would lead to a lot of confusion.
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100=w^2
W=10m/s
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What is the moment of inertia of a rotating system whose kinetic energy is 500 Joules when it is rotating at 40 radians / second?
500J=1/2I(40m/s)
.625kgm^2
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What definitions were required to answer these questions?
Ke=1/2mv^2, 1/2I(w)^2
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`q004. What is the moment of inertia of a styrofoam wheel in which 12 bolts, each of mass 100 grams, are inserted around a circle 50 cm in diameter and concentric with the axis of rotation, with another 8 bolts, each of mass 150 grams, inserted around a concentric circle 30 cm in diameter? Ignore the moment of inertia of the styrofoam itself.
1.2kg(.25m)^2
I=.075kg/m^2
And
1.2kg(.15m)^2
I=.027 kg/m^2
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The system has only one moment of inertia, which you obtain by adding the moments of inertia of all the individual masses.
So you would add the two results you got here.
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`q005. A long metal spring exerts forces of 500 Newtons, 400 Newtons, 250 Newtons, 100 Newtons, and 5 Newtons at respective lengths 20 meters, 18 meters, 16 meters, 13 meters and 10 meters.
What is its average tension between the 16 meter length and the 13 meter length?
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How much work would be required to stretch the cord from the 13 meter length to the 16 meter length?
About 179N
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That's a good estimate of the average force, but doesn't yet address the question of work.
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If the tension is conservative, how much potential energy would be lost as the cord's length decreases from 16 meters to 13 meters?
W=179x3
W=537J
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Good.
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Assuming the tension to be conservative, how much potential energy is built on each of the four intervals, as the spring is stretched from its 10 meter length to its 20 meter length?
179Nx10M
=-1790J
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The 179 N estimate was for a single 3-meter interval.
To answer the question you need to calculate the work done on each of the given intervals, use to find the potential energy change for each, and add your results.
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If all that potential energy could be converted to the kinetic energy of a rotating system whose moment of inertia is 2 kg m^2, what would be the angular velocity of that system?
1790=1/2(2kg/m^2)(w^2)
W=42.3 m/s
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What definitions are required to answer these questions?
KE, PE W
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`q006. How long would it take a freely falling object, starting from rest, to reach a speed of 300 meters / second, and how far would it travel during this time?
dT=2s dS=300m/s
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In two seconds a freely falling object starting from rest would achieve a velocity of 2 s * 9.8 m/s^2 = 19.6 m/s.
In other words, the object picks up 9.8 m/s every second (which is what it means to accelerate at 9.8 m/s^2), so in 2 s it picks up twice that or 19.6 m/s.
In 2 seconds, therefore, the velocity is nowhere near 300 m/s.
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`ds could not be 300 m/s . `ds is a displacement, measured in units of length (e.g., cm or meters or feet or miles etc.), while 300 m/s is a velocity, measured in units of length / time.
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What would be the change in the gravitational potential energy of a 100 kg mass falling in the manner described here?
Vo=0
Vf=300
Vave=150m/s
W=fxds
100kg75m/s^2
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Your instinct to multiply mass by acceleration to get force is good.
However the acceleration of a freely falling object is 9.8 m/s^2.
Nothing here is accelerating at 75 m/s^2.
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F=7500
W=-2,250
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There are no units on this quantity, and you haven't shown how it was calculated, so it's not clear what it means.
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PE= -2,250,000 J
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Unclear how this was found.
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If due to air resistance a 100 kg mass, falling from rest, requires 40 seconds to reach a speed of 300 meters / second, then what is your best estimate of the average force of air resistance?
100kg (300m/s/40s)
Fave=750 N
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Assuming uniform acceleration, (300 m/s) / (40 s) would be a valid calculation for the acceleration, and multiplying this by the 100 kg mass would give you the net force on that mass.
However that's the net force, not the force of air resistance.
What other forces act, and what do you therefore conclude to be the force of air resistance?
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You're making progress. My notes should help you clarify more of these ideas and procedures.
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