course Phy 201 `questionNumber 230000 `q001. Note that this assignment contains 3 questions.
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RESPONSE --> 50 cm of chain is hanging over the edge, and for each centimeter of length the chain has a mass of 15 g. So, the portion of the chain hanging over the edge has mass 50*15=750g (or .75 kg). To find the force of gravity acting on the chain that hangs over the edge, we will multiply the mass by the acceleration. .75kg*9.8m/s^2=7.35N The rest of the chain has length of 150 cm. 150 cm * 15 kg/cm=2250g or 2.25 kg 2.25kg*9.8m/s^2=22.05N We were told that this force has a .1 friction acting against it, meaning that 22.05*.1=2.2N frictional force So, the hanging force onthe change is 7.35N and the friction force is 2.2 N. Obviously, the friction force is acting opposite the hanging force and hence will be negative whereas the hanging force of gravity is positve. 7.35N-2.2N=5.15N (net force) Now we know the net force of the chain, so all we need to find is the total mass in order to be able to calculate the acceleration of the whole chain. 200cm of chain * 15kg/cm=3000g=3kg for the total mass a=Fnet/m =5.15N/3kg=1.72 cm/s^2 confidence assessment: 3
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20:04:13 `questionNumber 230000 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 --> I understand. self critique assessment: 3
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20:05:07 `questionNumber 230000 `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 am really not sure how to approach this one... confidence assessment: 0
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20:18:37 `questionNumber 230000 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 --> I will explain this in my own words: The frictional force has to equal the gravitational force for the chain to not accelerate. Because we have to find the mass of the chain by multiplying the centimeters by the kg/cm but we dont know the length in cm, x represents length but 15kg/cm is constant, as is the acceleration by which the mass is divided. Since there is still a variable, though, the equation remains x*.015kg/9.8cm/s^2. We can go ahead and multiply through the last two numbers, leaving the variable x*.147N/cm. Since we know the wholechain is 200cm in length, we know that 200-x (the unknown length required to keep the chain balanced and not accelerating). We treat this portion of the equation the same way we did the other, except that the forces of both gravity and friction are acting on this part of the chain. So, 200-x*.015kg/cm*9.8cm/s^2=.147N/cm*(200 cm-x). The frictional force is .1 of the gravitational newton force, so .1* .147N/cm*(200 cm-x)=.0147N/cm*200cm-x Since the forces have to be equal, we can set the two equations with variables equal to each other, since they each represent the forces exerted on either part of the chain. x*.147N/cm=.0147N/cm*200cm-x We follow the order of operations until we can solve directly for x. Then, we do so, yielding about 18 cm of length that can hang over for the system to not be accelerating. self critique assessment: 2
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20:25:33 `questionNumber 230000 `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 --> The object will encounter gravity, which will push downward, and air resistance, which will work to counter gravity. We know the object is mass 5kg, so we can multiply this by the acceleration of gravity and get the total force of gravity on the object: 5kg*9.8cm/s^2=49N Now, we literally use the equation given. We fill in 49 N for the force and then solve for v^2. 49N=.125v^2 392=v^2 19.8m/s=v confidence assessment: 3
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20:25:40 `questionNumber 230000 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 --> I understand. self critique assessment: 3
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