course phy121 ›ÖÉl–[×Év• µþäÒïC’…àýassignment #015
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18:31:24 Set 4 probs 1-7 If we know the net force acting on an object and the time during which the force acts, we can find the change in what important quantity?
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RESPONSE --> Fnet * `dt = impulse = momentum confidence assessment: 3
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18:31:29 ** You can find the change in the momentum. Fnet * `ds is change in KE; Fnet * `dt is change in momentum. **
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RESPONSE --> self critique assessment: 3
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18:31:45 What is the definition of the momentum of an object?
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RESPONSE --> Momentum = m * `dv confidence assessment: 3
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18:31:49 ** momentum = mass * velocity. Change in momentum is mass * change in velocity (assuming constant mass). UNIVERSITY PHYSICS NOTE: If mass is not constant then change in momentum is change in m v, which by the product rule changes at rate dp = m dv + v dm. If mass is constant `dm = 0 and dp = m dv so `dp = m * `dv. **
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RESPONSE --> self critique assessment:
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18:34:02 How do you find the change in the momentum of an object during a given time interval if you know the average force acting on the object during that time interval?
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RESPONSE --> Fave = `dp / `dt so `dp = Fave * dt confidence assessment: 3
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18:34:07 ** Since impulse = ave force * `dt = change in momentum, we multiply ave force * `dt to get change in momentum. **
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RESPONSE --> self critique assessment: 3
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18:34:41 How is the impulse-momentum theorem obtained from the equations of uniformly accelerated motion and Newton's Second Law?
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RESPONSE --> I do not know. confidence assessment: 0
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18:35:24 ** First from F=ma we understand that a=F/m. Now if we take the equation of uniformly accelerated motion vf= v0 + a'dt and subtract v0 we get vf-v0 = a'dt. Since vf-v0 = 'dv, this becomes 'dv = a'dt. Now substituting a=F/m , we get 'dv = (F/m)'dt Multiplying both sides by m, m'dv = F'dt **
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RESPONSE --> ok, I've written this down and will review it. self critique assessment: 3
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18:36:38 If you know the (constant) mass and the initial and final velocities of an object, as well as the time required to change from the initial to final velocity, there are two strategies we can use to find the average force exerted on the object. What are these strategies?
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RESPONSE --> one is calculating the initial and final velocities and dividing the difference in them by the time interval to get the Fave. I don't see the other looking at my notes. confidence assessment: 2
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18:37:38 ** The impulse-momentum theorem for constant masses is m `dv = Fave `dt. Thus Fave = m `dv / `dt. We could alternatively find the average acceleration aAve = (vf - v0) / `dt, which we then multiply by the constant mass to get Fave. **
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RESPONSE --> self critique assessment:
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18:37:52 ** The impulse-momentum theorem for constant masses is m `dv = Fave `dt. Thus Fave = m `dv / `dt. We could alternatively find the average acceleration aAve = (vf - v0) / `dt, which we then multiply by the constant mass to get Fave. **
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RESPONSE --> I see. self critique assessment: 2
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18:39:12 Class notes #14. How do we combine Newton's Second Law with an equation of motion to obtain the definition of energy?
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RESPONSE --> confidence assessment: 0
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18:46:18 ** a = F / m. vf^2 = v0^2 + 2 a `ds. So vf^2 = v0^2 + 2 (Fnet / m) `ds. Multiply by m/2 to get 1/2 mvf^2 = 1/2 m v0^2 + Fnet `ds so Fnet `ds = 1/2 m vf^2 - 1/2 m v0^2--i.e., work = change in KE. **
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RESPONSE --> I see the progression through these steps. I'm very afraid that I'm going to fail this class because of the amount of formulas. I'm not learning this and I'm having problems connecting this stuff together. I just need a C to transfer so I can graduate this December, but at this point I don't see myself being able to pull through this. I don't know what questions to ask to get help and I'm pretty lost at this point. Do not withdraw me, as I'm going to try to finish and pass. I'm not a good studier and never have been, and when I don't know what I have to study it makes matters worse. Plus having to watch my daughter while I'm working on this makes it frustraiting and hard to handle sometimes. I'm honestly trying, but I do much better solving problems with equasions rather than just being presented with equasions and trying to memorize them. Any insight or help you could provide would be great, as I'm really getting discouraged. Thanks! self critique assessment: 0
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18:48:05 What is kinetic energy and how does it arise naturally in the process described in the previous question?
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RESPONSE --> KE is the energy a moving object can exert on another object. It arises as its PE is used up. confidence assessment: 2
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18:48:21 ** KE is the quantity 1/2 m v^2, whose change was seen in the previous question to be equal to the work done by the net force. **
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RESPONSE --> self critique assessment: 0
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18:49:02 What forces act on an object as it is sliding up an incline?
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RESPONSE --> What ever force is moving it, gravity, and friction all do work on the object. confidence assessment: 3
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18:49:23 ** Gravitational force can be broken into two components, one parallel and one perpendicular to the ramp. The normal force exerted by the ramp is an elastic force, and unless the ramp breaks the normal force is equal and opposite to the perpendicular component of the gravitational force. Frictional force arises from the normal force between the two surfaces, and act in the direction opposed to motion. The gravitational force is conservative; all other forces in the direction of motion are nonconservative. COMMON ERROR: The Normal Force is in the upward direction and balances the gravitational force. COMMENT: The normal force is directed only perpendicular to the incline and is in the upward direction only if the incline is horizontal. The normal force cannot balance the gravitational force if the incline isn't horizontal. Friction provides a component parallel to the incline and opposite to the direction of motion. **
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RESPONSE --> self critique assessment: 0
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18:51:27 For an object sliding a known distance along an incline how do we calculate the work done on the object by gravity? How do we calculate the work done by the object against gravity?
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RESPONSE --> The work done by gravity will be the Fnet * `ds in the vertical direction. The work done by the object against gravity will be the opposite of the previous quantity. confidence assessment: 3
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18:51:43 ** The gravitational force is m * g directly downward, where g is the acceleration of gravity. m * g is the weight of the object. If we know change in vertical position then we can simply multiply weight m * g with the vertical displacement `dy, being careful to keep track of which is positive and/or negative. Alternatively it is instructive to consider the forces in the actual direction of motion along the incline. For small inclines the component of the gravitational force which is parallel to the incline is approximately equal to the product of the weight and the slope of the incline, as seen in experiments. The precise value of the component parallel to the incline, valid for small as well as large displacements, is m g * sin(theta), where theta is the angle of the incline with horizontal. This force acts down the incline. If the displacement along the incline is `ds, measured with respect to the downward direction, then the work done by gravity is the product of force and displacement, m g sin(theta) * `ds. If `ds is down the incline the gravitational component along the incline is in the same direction as the displacement and the work done by gravity on the system is positive and, in the absence of other forces in this direction, the KE of the object will increase. This behavior is consistent with our experience of objects moving freely down inclines. If the displacement is upward along the incline then `ds is in the opposite direction to the gravitational force and the work done by gravity is negative. In the absence of other forces in the direction of the incline this will result in a loss of KE, consistent with our experience of objects coasting up inclines. The work done against gravity is the negative of the work done by gravity, positive for an object moving up an incline (we have to use energy to get up the incline) and negative for an object moving down the incline (the object tends to pick up energy rather than expending it) **
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RESPONSE --> self critique assessment: 0
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18:53:28 For an object sliding a known distance along an incline how do we calculate the work done by the object against friction? How does the work done by the net force differ from that done by gravity?
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RESPONSE --> Work done by the object against gravity will the the product of the `ds and Fgrav. The work done by Fnet is the force that has overcome the Fgrav and is moving the object. When you add them together, you get the total force applied to the object. confidence assessment: 3
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18:53:45 ** The work done against friction is the product of the distance moved and the frictional force. Since the force exerted by friction is always opposed to the direction of motion, the force exerted by the system against friction is always in the direction of motion so the work done against friction is positive. The net force on the system is sum of the gravitational component parallel to the incline and the frictional force. The work done by the net force is therefore equal to the work done by gravity plus the work done by the frictional force (in the case of an object moving up an incline, both gravity and friction do negative work so that the object must do positive work to overcome both forces; in the case of an object moving down an incline gravity does positive work on the system while friction, as always, does negative work on the system; in the latter case depending on whether the work done by gravity on the system is greater or less than the frictional work done against the system the net work done on the system may be positive or negative) **
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RESPONSE --> self critique assessment: 0
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18:54:11 Explain why the restoring force on a simple pendulum is in nearly the same proportion to the weight of the pendulum as its displacement from equilibrium to its length, and explain which assumption is made that makes this relationship valid only for displacements which are small compared to pendulum length.
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RESPONSE --> I do not know. confidence assessment: 0
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18:56:19 ** In terms of similar triangles: The reason the approximation only works for small displacements is because the sides used on one triangle are not the same as the sides used on the other. From the triangle we see that the restoring force and the weight are at right angles, while the length and horizontal displacement of the pendulum from equilibrium are the hypotenuse and the horizontal leg of a triangle and hence are not at right angles. For small angles the two long sides of the triangle are approximately equal so the discrepancy doesn't make much difference. For larger angles where the two long sides are significantly different in length, the approximation no longer works so well. In terms of components of the vectors: The tension force is in the direction of the string. The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the total tension as the length of the pendulum to the horizontal displacement (just draw the picture). The vertical component of the tension force must be equal to the weight of the pendulum, since the pendulum is in equilibrium. If the displacement is small compared to the length the vertical component of the tension force will be very nearly equal to the tension force. So the previous statement that 'The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the total tension as the length of the pendulum to the horizontal displacement' can be replaced by the statement that 'The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the weight of the pendulum as the length of the pendulum to the horizontal displacement. **
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RESPONSE --> self critique assessment:
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18:57:06 prin and gen phy: 6.4: work to push 160 kg crate 10.3 m, horiz, no accel, mu = .50.
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RESPONSE --> confidence assessment:
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18:58:34 The net force on the crate must be zero, since it is not accelerating. The gravitational force on the crate is 160 kg * 9.8 m/s^2 = 1570 N, approx. The only other vertical force is the normal force, which must therefore be equal and opposite to the gravitational force. As it slides across the floor the crate experiences a frictional force, opposite its direction of motion, which is equal to mu * normal force, or .50 * 1570 N = 780 N, approx.. The only other horizontal force is exerted by the movers, and since the net force on the crate is zero the movers must be exerting a force of 780 N in the direction of motion. The work the movers do in 10.3 m is therefore work = Fnet * `ds = 780 N * 10.3 m = 8000 N m = 8000 J, approx..
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RESPONSE --> I see. self critique assessment: 3
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