course Phy 121
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09:39:37 We put energy into the rubber-band-and-rail system, in the form of the work we do stretching the rubber band. In detail, what happens to this energy from the instant we start pulling back to the instant the rail stops?
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RESPONSE --> Most of the energy of the elastic is stored as PE of the rubber band. When the system is released, the PE decreases and goes directly into the KE of the system.
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09:39:41 The system stores most of the energy as elastic PE of the rubber band. When the system is released the PE decreases and goes into the KE of the system. There are also thermodynamic effects. As the rubber band is stretched the thermal energy that results from the stretching is released into the air. When the rubber band becomes unstretched it cools off, but the thermal energy that was released does not return to the rubber band.
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RESPONSE --> ok
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09:41:18 What happens to the KE of the rail as it slides across the floor? Where does it go?
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RESPONSE --> The KE is lost to friction as it slides across the floor. The result of friction is from thermal energy, or heat. It simply goes away as it slides.
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09:41:21 ** The KE was gained at the expense of the rubber band's PE, but as it slides across the floor it is lost to friction. The result of friction is thermal energy, or heat, which is dissipated from the rail and the surface. **
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RESPONSE --> ok
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09:41:44 From the point of view of the rail, is it doing positive work or negative work as it slides across the floor?
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RESPONSE --> It's doing positive work.
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09:41:47 ** It's doing positive work, but the reason is that it is exerting a force against friction, and this force is in the direction of its motion. **
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RESPONSE --> ok
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09:43:57 Does the rubber band supply more or less energy to the rail than the energy we put into it, and where does the difference go?
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RESPONSE --> It puts more energy in than we do, but it dissapates as it slides across the floor due to friction, and continues until all the energy is dissipated and the rail stops.
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09:44:00 When we pull the rail back, we do work, most (but not all) of which will be recovered after we release the rail. This work will therefore be present in the rail in the form of energy of motion. Before we release the rail, this work is potentially there; the potential will become reality after we release the rail. At the instant the rail reaches its original position the rubber band ceases to accelerate it, and it has its maximum energy of motion. As the rail slides across the floor, this energy is dissipated in the form of work done against friction. This continues until all the energy is dissipated and the rail stops.
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RESPONSE --> ok
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09:44:59 Does the rubber band exert more or less average force when it accelerates the rail than when it was pulled back?
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RESPONSE --> Less average force. It can't exert more than we put into the system.
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09:45:02 ** It can't exert more average force because if it did it would release more energy than was put into it. **
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RESPONSE --> ok
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09:45:47 Is a cooler rubber band stiffer or less stiff than a warmer one?
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RESPONSE --> The cooler rubber band is less stiff than the warmer one.
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09:45:51 The cooler rubber band is less stiff than the warmer one, contrary to our intuition about how things stiffen when they are cooled.
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RESPONSE --> cool
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ͣƥͫ^j assignment #013 Wq}PsӲqʧp Physics I Class Notes 02-24-2006
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09:48:02 What would a graph of potential energy vs. stretch look like for the rubber band?
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RESPONSE --> We know potential energy increases with stretch. We also know the force required to stretch the rubber band increases with the size of the stretch. This means the PE increases at an increasing rate.
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09:48:05 * potential energy increases with stretch; and since the force required to stretch the rubber band increases with stretch, the PE increases at an increasing rate with respect to the stretch. **
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RESPONSE --> ok
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09:49:35 What would a graph of kinetic energy vs. distance look like for the rail sliding across the floor? Would the graph be linear or would it curve? If it would curve, in what direction would it curve?
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RESPONSE --> The graph would be linear and decrease at a decreasing rate.
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09:49:39 ** since the rail is moving more and more slowly, and since the work done against friction depends on the distance moved, the rail will lose KE more and more slowly. So the graph of KE vs. distance will decrease, but at a decreasing rate. **
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RESPONSE --> ok
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09:51:09 How is it that we can regard the force of gravity as equivalent to two forces, one parallel and one perpendicular to the ramp?
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RESPONSE --> We regard them as being equal.
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09:51:12 The forces wParallel and wPerpendicular are to be regarded as completely equivalent to the weight. We should understand that if we pulled on an object with two forces equal to these, and in the indicated directions, they could have an effect identical to that of the downward force of gravity. So the force of gravity could be replaced by these two forces. Given a high enough coefficient of friction, the force acting parallel to the incline could be equal and opposite to the frictional force. The normal force is the elastic force required to exactly balance the weight component perpendicular to the ramp.
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RESPONSE --> ok
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09:51:17 ""
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RESPONSE --> ok
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HƜiЫw assignment #014 Wq}PsӲqʧp Physics I Class Notes 02-24-2006
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09:55:03 How does the velocity vs. clock time trapezoid give us the two basic equations of motion?
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RESPONSE --> The displacement between any two clock times is equal to the area of the trapezoid between the two clock times, and that gives us: `ds = (v0 + vf) / 2 * `dt. The slope of the graph will be the acceleration a, and the change in velocity between two clock times. So we get: vf = v0 + a `dt.
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09:55:06 The displacement between any two clock times is equal to the area of the trapezoid between these two clock times. This area is equal to the product of the average altitude of the trapezoid and its width, which gives us the first basic equation of uniformly accelerated motion: `ds = (v0 + vf) / 2 * `dt. Under the same conditions the slope of the graph will be the acceleration a, and the change in velocity between two clock times will be represented by the rise of a slope triangle who slope is a and whose run is the time interval `dt between the two clock times. The velocity change will therefore be `dv = a * `dt, and the final velocity will be equal to the initial velocity plus this change: vf = v0 + a `dt. This is the second basic equation of motion.
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RESPONSE --> ok
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09:56:41 When a wound-up friction car is released on the level surface, what do we see as a result of the potential energy conversion?
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RESPONSE --> When it is released, most of this energy is converted to kinetic energy, and there is friction loss as well.
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09:56:45 Most of the work done to wind the spring goes into the elastic potential energy of the spring. When it is released, most of this energy is then converted to kinetic energy. In each part of the process there is some friction loss in the mechanism.
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RESPONSE --> ok
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09:58:34 When a wound-of friction car is released up a ramp, what to we see as a result of the potential energy conversion?
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RESPONSE --> As the car speeds up, this increases its KE. A little bit of the energy is dissipated in the form of thermal energy due to friction on the car.
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09:58:37 The car climbs the ramp, increasing its gravitational PE. The car also speeds up, increasing its KE. Some of the energy is dissipated in the form of thermal energy as a result of friction.
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RESPONSE --> ok
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09:59:53 What forces act on an object sliding up or down an incline?
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RESPONSE --> Gravity and friction.
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09:59:57 A normal force will be exerted by the ramp. Gravity acts on the object; usually the resulting weight is expressed as two components, one parallel and one perpendicular to the incline. There is also a frictional force acting in the direction opposite to the motion of the object.
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RESPONSE --> ok
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10:02:07 What energy changes take place as an object slides up or down an incline?
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RESPONSE --> Change in Pe and KE as it moves up the ramp.
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10:02:11 The parallel component of the weight is wParallel = w sin(`theta), and to slide the object through displacement `ds at a constant velocity requires a force equal and opposite to this component. As a result work `dW = w sin(`theta) * `ds, if the positive direction is chosen as up the incline. If we simply raise the object vertically through a distance equal to its vertical rise, we will have to exert a force equal in magnitude to its weight. The vertical distance through which we lift the object will be `dy = `ds sin(`theta), so the work done will be `dW = w * `dy = w * (`ds sin(`theta)). It should be clear that this is the same work contribution found before.
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RESPONSE --> ok
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10:04:14 If we know the mass and length of a pendulum, how can we determine the force required to displaced pendulum a given small distance from equilibrium?
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RESPONSE --> If you know mass and length, then use F / mg = x / L. Rearranged, it looks like this, F = mg * (x / L), which is the weight of the pendulum multiplied by the ratio x / L of displacement to length.
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10:04:17 When a pendulum is displaced a certain distance from its equilibrium, the forces on the pendulum will be in equilibrium if the tension force directed along the pendulum string has a vertical component equal to the weight and a horizontal component equal to F. Therefore, we will have equilibrium if the ratio of F to the weight is the same as the ratio of the displacement x to the length L. That is, F / mg = x / L. It follows that F = mg * (x / L), which is the weight of the pendulum multiplied by the ratio x / L of displacement to length.
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RESPONSE --> ok
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hҗɣVɒ assignment #015 Wq}PsӲqʧp Physics I Class Notes 02-24-2006
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10:06:44 Why do we expect the velocity attained by a ball on a ramp to be proportional to the square root of the vertical position change of the ball?
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RESPONSE --> Because its gravitational potential energy will convert to kinetic energy.
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10:06:50 If the object was released from rest and and allowed to fall freely through a downward distance `dy equal to the vertical distance traveled on the ramp, its gravitational potential energy will convert to kinetic energy. In this case, setting the potential energy decrease equal to the kinetic energy increase (i.e., `dKE = -`dPE) gives.5 m v^2 = m g `dy. We solve to obtain v = `sqrt(2 g `dy). This demonstrates that v is proportional to`sqrt(`dy).
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RESPONSE --> oh okay
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10:09:07 Why do we expected distance traveled by a ball after being projected horizontally off of a ramp and falling a fixed distance to be proportional to the velocity with which the ball leaves the ramp?
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RESPONSE --> If v^2 is proportional to dy, then for a constant k, note v^2= k * dy. dx^2 is also proportional to dy, and thus proportional to the sqrt(dy). This means `dy is proportional to v^2, and that sqrt(`dy) is proportional to v, and that dx is proportional to the velocity v.
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10:09:10 If v^2 is proportional to dy, then for some constant k we have v^2 = k * dy. From experimental results it appears that dx^2 is also proportional to dy, so that dx is emprircally proportional to sqrt(dy). Since `dy is proportional to v^2, it follows that sqrt(`dy) is proportional to v, and we finally conclude that dx is proportional to the velocity v with which the ball is projected horizontally from the ramp.
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RESPONSE --> ok
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10:15:17 Why is the distance traveled by the projectile in this situation less than that predicted from the velocity v, where v is determined by setting 1/2 m v^2 equal to the potential energy loss on the ramp? Why is the distance still less even if frictional losses are taken into consideration?
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RESPONSE --> The ball gains kinetic energy in addition to its velocity associated with its rotational motion. As a ball rolls down a smooth ramp, its PE loss converted to rotational KE. Frictional losses also further reduce velocity.
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10:15:21 This situation is due mostly to the fact that in addition to attaining a velocity v the ball also rolls down the ramp and therefore gains kinetic energy associated with its rotational motion. As a ball rolls (without slipping) down a smooth ramp, it will turn out that 2/7 of its PE loss converted to rotational KE. This means that the KE associated with its velocity v on the ramp, which is called translational kinetic energy, theoretically, should only be 5/7 of the PE loss. Frictional losses further reduce v.
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RESPONSE --> ok
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