course Phy 201 dׁ䚭|yVassignment #023
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11:33:23 `q001. Note that this assignment contains 3 questions. . A chain 200 cm long has a density of 15 g/cm. Part of the chain lies on a tabletop, with which it has a coefficient of friction equal to .10. The other part of the chain hangs over the edge of the tabletop. If 50 cm of chain hang over the edge of the tabletop, what will be the acceleration of the chain?
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RESPONSE --> 200cm-50cm=150cm on the tabletop F=ma Density=mass/volume 50cm*15g/cm=750g=.750kg .750kg*9.8m/s^2=7.35kgm/s^2=7.35N 200cm*15g/cm=3000g=3kg 7.35N=3kg*a Divide by 3 kg 2.45m/s^2=a confidence assessment: 1
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11:37:40 The part of the chain hanging over the edge of the table will experience an unbalanced force from gravity and will therefore tend to accelerate chain in the direction of the hanging portion. The remainder of the chain will also experience the gravitational force, but this force will be countered by the upward force exerted on the chain by the table. The force between the chain and the table will give rise to a frictional force which will resist motion toward the hanging portion of the chain. If 50 cm of chain hang over the edge of the tabletop, then we have 50 cm * (15 g/cm) = 750 grams = .75 kg of chain hanging over the edge. Gravity will exert a force of 9.8 meters/second ^ 2 * .75 kg = 7.3 Newtons of force on this mass, and this force will tend to accelerate the chain. The remaining 150 cm of chain lie on the tabletop. This portion of the chain has a mass which is easily shown to be 2.25 kg, so gravity exerts a force of approximately 21 Newtons on this portion of the chain. The tabletop pushes backup with a 21 Newton force, and this force between tabletop and chain results in a frictional force of .10 * 21 Newtons = 2.1 Newtons. We thus have the 7.3 Newton gravitational force on the hanging portion of the chain, resisted by the 2.1 Newton force of friction to give is a net force of 5.2 Newtons. Since the chain has a total mass of 3 kg, this net force results in an acceleration of 5.2 N / (3 kg) = 1.7 meters/second ^ 2, approximately.
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RESPONSE --> 150cm on the tabletop * 15g/cm=2.25kg 2.25kg*9.8m/s^2=21N Multiply the force on the tabletop by the frictional force .10*21N=2.1N Should use the 7.3N minus the frictional force of 2.1N to get net force of 5.2N 5.2N/3kg=1.7m/s^2 self critique assessment: 2
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12:00:06 `q002. What is the maximum length of chain that can hang over the edge before the chain begins accelerating?
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RESPONSE --> I calculated the forces for overhanging lengths of 10 and 20cm and noticed that for 20cm there would be acceleration but not for 10cm. Next I plugged in 15cm of overhang but the force of friction was larger than the force of overhang. I went to 19 cm and the frictional force was less than on the overhang, indicating acceleration. I used 18cm which seemed to allow for the least difference between the forces on the overhang and the table portion. confidence assessment: 1
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12:07:38 The maximum length that can hang over is the length for which the frictional force opposing motion is precisely equal to the gravitational force on the hanging portion of the chain. If x stands for the length in cm of the portion of chain hanging over the edge of the table, then the mass of the length is x * .015 kg / cm and it experiences a gravitational force of (x * .015 kg / cm) * 9.8 m/s^2 = x * .147 N / cm. The portion of chain remaining on the tabletop is 200 cm - x. The mass of this portion is (200 cm - x) * .015 kg / cm and gravity exerts a force of (200 cm - x) * .015 kg / cm * 9.8 meters/second ^ 2 = .147 N / cm * (200 cm - x) on this portion. This will result in a frictional force of .10 * .147 N / cm * (200 cm - x) = .0147 N / cm * (200 cm - x). Since the maximum length that can hang over is the length for which the frictional force opposing motion is precisely equal to the gravitational force on the hanging portion of the chain, we set the to forces equal and solve for x. Our equation is .0147 N / cm * (200 cm - x) = .147 N/cm * x. Dividing both sides by .0147 N/cm we obtain 200 cm - x = 10 * x. Adding x to both sides we obtain 200 cm = 11 x so that x = 200 cm / 11 = 18 cm, approx..
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RESPONSE --> x=length of hanging portion mass=x*.015kg/cm gravitational force=x*(.015kg/cm) *9.8m/s^2=x*.147N/cm Tabletop portion= 200cm-x mass=(200cm-x)*.015kg/cm gravitational force= (200cm-x)*.015kg/cm *9.8m/s^2=.147N/cm *(200cm-x) Frictional force=.10*.147N/cm *(200cm-x)=.0147N/cm *(200cm-x) .0147N/c, *(200cm-x)=.147N/cm *x 200cm-x=10*x 200cm=11x x=200cm/11=18cm self critique assessment: 2
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12:16:18 `q003. The air resistance encountered by a certain falling object of mass 5 kg is given in Newtons by the formula F = .125 v^2, where the force F is in Newtons when the velocity v is in meters/second. As the object falls its velocity increases, and keeps increasing as it approaches its terminal velocity at which the net force on the falling object is zero, which by Newton's Second Law results in zero acceleration and hence in constant velocity. What is the terminal velocity of this object?
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RESPONSE --> mass=5kg Force=5kg*9.8m/s^2=49N 49N=.125v^2 Divide by .125 392m^2/s^2=v^2 +-19.80=v confidence assessment: 1
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12:17:57 Only two forces act on this object, the downward force exerted on it by gravity and the upward force exerted by air resistance. The downward force exerted by gravity remains constant at 5 kg * 9.8 meters/second ^ 2 = 49 Newtons. When this force is equal to the .125 v^2 Newton force of friction the object will be at terminal velocity. Setting .125 v^2 Newtons = 49 Newtons, we divide both sides by .125 Newtons to obtain v^2 = 49 Newtons/(.125 Newtons) = 392. Taking square roots we obtain v = `sqrt (392) = 19.8, which represents 19.8 meters/second.
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RESPONSE --> 5kg*9.8m/s^2=49N .125v^2=49N v^2=49N/.125N=392 v=19.8m/s self critique assessment: 2
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