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13:01:05 Query gen phy problem 4.07 force to accelerate 7 g pellet to 125 m/s in .7 m barrel
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RESPONSE --> 7g = .007kg a = (vf^2 - v0^2) / (2*'ds) = (125m/s)^2 / (2*.7m) a = 11160.7m/s^2
F = m*a = .007kg * 11160.7m/s^2 = 78.125N.................................................
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13:02:42 ** The initial velocity of the bullet is zero and the final velocity is 175 m/s. If we assume uniform acceleration (not necessarily the case but not a bad first approximation) the average velocity is (0 + 125 m/s) / 2 = 62.5 m/s and the time required for the trip down the barrel is .7 m / (62.5 m/s) = .011 sec, approx..
Acceleration is therefore rate of velocity change = `dv / `dt = (125 m/s - 0 m/s) / (.11 sec) = 11000 m/s^2, approx.. The force on the bullet is therefore F = m a = .007 kg * 11000 m/s^2 = 77 N approx. **......!!!!!!!!...................................
RESPONSE --> ok.
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13:02:47 univ phy *&*& now a parachuist *&*&problem 4.38 (4.34 10th edition) fish on balance, reading 60 N when accel is 2.45 m/s^2 up; find true weight, situation when balance reads 30 N, balance reading when cable breaks
What is the net force on the fish when the balance reads 60 N? What is the true weight of the fish, under what circumstances will the balance read 30 N, and what will the balance read when the cable breaks?......!!!!!!!!...................................
RESPONSE --> ok
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13:02:50 ** Weight is force exerted by gravity.
Net force is Fnet = m * a. The forces acting on the fish are the 50 N upward force exerted by the cable and the downward force m g exerted by gravity. So m a = 50 N - m g, which we solve for m to get m = 50 N / (a + g) = 50 N / (2.45 m/s^2 + 9.8 m/s^2) = 50 N / 12.25 m/s^2 = 4 kg. If the balance reads 30 N then Fnet is still m * a and we have m a = 30 N - m g = 30 N - 4 kg * 9.8 m/s^2 = -9.2 N so a = -9.2 N / (4 kg) = -2.3 m/s^2; i.e., the elevator is accelerating downward at 2.3 m/s^2. If the cable breaks then the fish and everything else in the elevator will accelerate downward at 9.8 m/s^2. Net force will be -m g; net force is also Fbalance - m g. So -m g = Fbalance - m g and we conclude that the balance exerts no force. So it reads 0. **......!!!!!!!!...................................
RESPONSE --> ok
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13:04:49 STUDENT QUESTION: I had trouble with the problems involving tension in lines. For example the Fish prob.
Prob#9 A person yanks a fish out of water at 4.5 m/s^2 acceleration. His line is rated at 22 Newtons Max, His line breaks, What is the mass of the fish. Here's what I did. Sum of F = Fup + F down -22 N = 4.5 m/s^2 * m(fish) - 9.8 m/s^2 * m(fish) -22N = -5.3 m/s^2 m(fish) m(fish) = 4.2 kg I know its wrong, I just don't know what to do.I had the same problem with the elevator tension on problem 17.......!!!!!!!!...................................
RESPONSE --> I don't know if I'm right either, but I just did the F = m*a equations, where m = F / a = 22N / 4.5m/s^2 m = 4.89kg...
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13:06:48 ** Think in terms of net force.
The net force on the fish must be Fnet = m a = m * 4.5 m/s^2. Net force is tension + weight = T - m g, assuming the upward direction is positive. So T - m g = m a and T = m a + m g. Factoring out m we have T = m ( a + g ) so that m = T / (a + g) = 22 N / (4.5 m/s^2 + 9.8 m/s^2) = 22 N / (14.3 m/s^2) = 1.8 kg, approx.. The same principles apply with the elevator. **......!!!!!!!!...................................
RESPONSE --> Ok this makes sense.
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15:31:08 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 --> We lose energy as we work to pull back the rubberband. Then when it's released, the energy (minus the friction or other forces involved) is returned to us.
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15:31:31 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.......!!!!!!!!...................................
RESPONSE --> ok
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15:33:23 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 of the rail decreases as it slides across the floor. I'm not sure where it goes.
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15:33:50 ** 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 so it's simply lost due to the friction
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15:34:58 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 --> I think it is doing negative work... I say this because I think the rail does positive work on the rubberband when it's being stretched back, therefore it would be doing negative work as the rubberband pushes it across the floor.
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15:35:48 ** 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 --> so it has nothing to do with the rubberband. it's positive because it's going against friction in the direction of motion.
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15:37:19 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 --> I would say that the rubberband supplies the same amount as energy as we put into it, but I'm thinking that because of thermal energy, then some of the energy we put into the rubberband will be lost.
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15:42:06 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 --> k
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15:42:37 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 --> My guess is more average force.
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15:42:47 ** 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 --> oh i see.
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15:43:52 Is a cooler rubber band stiffer or less stiff than a warmer one?
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RESPONSE --> stiffer i would think.
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15:44:35 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 --> oh. i figured since the heat would cause the molecules to move and bounce off each other at a higher rate, then the rubber band would be more flexible.
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15:53:45 What would a graph of potential energy vs. stretch look like for the rubber band?
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RESPONSE --> Potential energy would be on the y-axis and stretch would be on the x-axis and I think the line would be increasing at a pretty constant rate.
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15:53:57 * 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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15:56:06 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 --> I think the graph would curve upward since more KE is needed to push the rail the further distances.
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15:57:24 ** 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. **......!!!!!!!!...................................
RESPONSE --> Ok so even though more energy is needed to push the rail further, more and more KE is lost due to friction.
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15:58:33 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 --> You have gravity that is pushing down on the object while it's on the ramp, which is perpendicular, then there's the gravity that either works for or against the object as it coasts up or down the ramp.
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16:01:36 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.......!!!!!!!!...................................
RESPONSE --> ok
assignment #014 ~ȪsJ콥 Physics I Class Notes 10-22-2005......!!!!!!!!...................................
16:10:16 How does the velocity vs. clock time trapezoid give us the two basic equations of motion?
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RESPONSE --> By looking at the trapezoid, the difference between the velocities can be averaged for the average altitude of the trapezoid. Then multiplying the vAve by the time interval between the two points gives us the distance. This means that the area of the trapezoid is what can be used to determine 'ds = vAve * 'dt.
By figuring out the difference in the velocities and dividing it by the time interval, we get acceleration, which is the slope of the graph (rise/run = 'dv / 'dt).................................................
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16:10:36 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.......!!!!!!!!...................................
RESPONSE --> ok
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16:15:49 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 --> If all the energy was converted from PE to KE, then it seems that the KE is lower than the initial potential energy put into it.
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16:15:59 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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16:18:30 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 --> The PE is converted to KE but the energy decreases due to the friction working against the object.
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16:20:46 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 --> If the car is speeding up while it goes up an incline, then it won't stop until friction causes it to, and I thought in order for it to stop, the energy has to be lost until it equals the forces... so how is it that the KE continues to increase?
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16:21:14 What forces act on an object sliding up or down an incline?
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RESPONSE --> perpendicular and parallel gravitational forces, friction, and thermal energy.
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16:21:26 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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16:23:23 What energy changes take place as an object slides up or down an incline?
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RESPONSE --> The Potential energy can be converted to kinetic energy... other than that, I'm not sure.
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16:24:33 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.......!!!!!!!!...................................
RESPONSE --> I read this in the notes, but the question asked for energy changes and I didn't know that this is what it was referring to.
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16:27:52 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 --> F / mg = x / L. The force - weight ratio has to equal the displacement x - Length ratio...
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16:28:27 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.......!!!!!!!!...................................
RESPONSE --> I didn't use the equation to solve for F, but it seems simple enough.
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assignment #015 ~ȪsJ콥 Physics I Class Notes 10-22-2005
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16:31:06 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 --> The KE attained on the ramp is proportional to `dy and to v^2, so the velocity v attained on the ramp is proportional to `sqrt(y). Since the horizontal velocity of the projectile is constant, the horizontal distance `dx it travels is proportional to v and hence to `sqrt(y). Since `dx is proportional to `sqrt(y), y is proportional to `dx^2.
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16:33:28 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).......!!!!!!!!...................................
RESPONSE --> ok
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16:34:55 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 the ball is then projected in the horizontal direction with the resulting velocity and allowed to fall freely through some distance `dh, the horizontal distance it moves will clearly be greater when the velocity with which it was projected is greater.
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16:39:21 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.......!!!!!!!!...................................
RESPONSE --> I wasn't looking in the right part of the notes for this question, which is why what I posted the past few questions doesn't match up at all. What is the constant k variable?
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16:44:39 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 --> In addition to attaining a velocity v the ball also rolls, and is a result gains a kinetic energy associated with its rotational motions. It turns out that, even in an ideal situation were there is no frictional loss, a ball rolling down a smooth ramp without slipping will have 2/7 of its potential energy loss converted to rotational kinetic energy.
This means that the kinetic energy associated with the velocity v, which is called the translational kinetic energy, should theoretically be only 5/7 of the potential energy loss. Friction will reduce observed translational kinetic energy even a bit more..................................................
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16:44:57 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 -->