course phy 201 ??????????????assignment #014??????j?????t??
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20:33:22 How does the velocity vs. clock time trapezoid give us the two basic equations of motion?
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RESPONSE -->
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W?z???m?????i?? assignment #014 014. `query 14 Physics I 03-25-2009
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20:42:51 set 3 intro prob sets If you calculate the acceleration on a mass m which starts from rest under the influence of a constant net force Fnet and multiply by a time interval `dt what do you get? How far does the object travel during this time and what velocity does it attain? What do you get when you multiply the net force by the distance traveled? What kinetic energy does the object attain?
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RESPONSE --> Fnet *ds = W F=ma a = F/m a * dt a = dv/dt you would get change in velocity v0 = 0 vf = v0 +at ds = v0*dt +.5a*dt^2 W = KE confidence assessment: 3
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20:45:52 **STUDENT ANSWER AND INSTRUCTOR COMMENTS: a*'dt = the final velocity if V0=0. to get the change in position you would divide the final velocity(since V0=0) by 2 to get the average velocity and then multiply that by the 'dt to get the units of distance traveled. Multiply that by the 'dt to get the units of distance traveled. It attains a Vf of a*'dt as shown above because V0=0, if V0 was not zero you would have to add that to the a*'dt to get the final velocity. When you multiply Fnet by 'dt you get the same thing you would get if you multiply the mass by the change in velocity(which in this case is the same as the final velocity). This is the change in momentum. The Kinetic Energy Attained is the forcenet multiplied by the change in time. a = Fnet / m. So a `dt = Fnet / m * `dt = vf. The object travels distance `ds = v0 `dt + .5 a `dt^2 = .5 Fnet / m * `dt^2. When we multiply Fnet * `ds you get Fnet * ( .5 Fnet / m * `dt^2) = .5 Fnet^2 `dt^2 / m. The KE attained is .5 m vf^2 = .5 m * ( Fnet / m * `dt)^2 = .5 Fnet^2 / m * `dt^2. Fnet * `ds is equal to the KE attained. The expression for the average velocity would be [ (v0 + a * `dt) + v0 ] / 2 = v0 + 1/2 a `dt so the displacement would be (v0 + 1/2 a `dt) * `dt = v0 `dt + 1/2 a `dt^2. This is equal to (v0 `dt + 1/2 a `dt^2) * Fnet = (v0 `dt + 1/2 a `dt^2) * m a , since Fnet = m a. **
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RESPONSE --> again it is benificial to see multiple methods to find different variables these equations are broken down more than I did but i fully understand the concepts self critique assessment: 2
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20:50:30 Define the relationship between the work done by a system against nonconservative forces, the work done against conservative forces and the change in the KE of the system. How does PE come into this relationship?
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RESPONSE --> W = fnet * ds and nonconservative forces Work done will not be stored therefore some or all will be lost ex heat, however PE and KE will be decreased work doen against conservative forces will recover stored energy and PE or KE will be increased PE and KE are linked because PE is the stored energy while KE is the energy in motion. confidence assessment: 2
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20:52:06 ** The work done by the system against all forces will decrease the KE by an equal amount. If some of the forces are conservative, then work done against them increases the PE and if PE later decreases this work will be recovered. Work done against non-conservative forces is not stored and cannot be recovered. STUDENT RESPONSE WITH INSTRUCTOR COMMENTARY: The work done by a system against nonconservative forces is the work done to overcome friction in a system- which means energy is dissipated in the form of thermal energy into the 'atmosphere.' Good. Friction is a nonconservative force. However there are other nonconservative forces--e.g., you could be exerting a force on the system using your muscles, and that force could be helping or hindering the system. A rocket engine would also be exerting a nonconservative force, as would just about any engine. These forces would be nonconservative since once the work is done it can't be recovered. STUDENT RESPONSE WITH INSTRUCTOR COMMENTS: The work done by a system against conservative forces is like the work to overcome the mass being pulled by gravity. INSTRUCTOR COMMENT: not bad; more generally work done against conservative force is work that is conserved and can later be recovered in the form of mechanical energy **
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RESPONSE --> self critique assessment:
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20:52:46 class notes: rubber band and rail How does the work done to stretch the rubber band compare to the work done by the rubber band on the rail, and how does the latter compare to the work done by the rail against friction from release of the rubber band to the rail coming to rest?
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RESPONSE --> i have not reviewed these notes confidence assessment: 1
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20:55:54 ** The work done to stretch the rubber band would in an ideal situation be available when the rubber band is released. Assuming that the only forces acting on the rail are friction and the force exerted by the rubber band, the work done by the rail against friction, up through the instant the rail stops, will equal the work done by the rubber band on the rail. Note that in reality there is some heating and cooling of the rubber band, so some of the energy gets lost and the rubber band ends up doing less work on the rail than the work required to stretch it. **
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RESPONSE --> work done to stretch is conserved with the loss of some thermal energy Fnet = Frubberband -Ffriction *ds = work done by the rail Fnet = Frubberband *ds = wrok by rubberband self critique assessment: 1
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20:57:42 Why should the distance traveled by the rail be proportional to the F * `ds total for the rubber band?
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RESPONSE --> because the work done by the rubberband is exerted on the rail and it is taking more force to move the rail than the rail to resist the rubberband confidence assessment: 2
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21:04:23 ** The F `ds total of the rail when it is accelerated by the rubber band is equal Fave `ds, which is equal to to m * aAve * `ds. Here aAve is the average acceleration of the rail by the rubber band. 2 aAve `ds = vf^2 - v0^2 by the fourth equation of motion. So the F `ds total is proportional to the change in v^2. The rail is then stopped by the frictional force f; since f `ds is equal to m * a * `ds, where a is the acceleration of the sliding rail, it follows that f `ds is also proportional to the change in v^2. Change in v^2 under the influence of the rubber band (rest to max vel) is equal and opposite to the change in v^2 while sliding against friction (max vel back to rest), so work f `ds done by friction must be equal and opposite to F `ds. This ignores the small work done by friction while the rubber band is accelerating the rail. **
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RESPONSE --> self critique assessment:
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21:07:38 gen phy A person of mass 66 kg crouches then jumps to a height of .8 meters. From the crouches position to the point where the person leaves the ground the distance is 20 cm. What average force is exerted over this 20-cm distance?
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RESPONSE --> m= 66kg ds = .8 + .2 = 1m Fnet =ma Fnet = 66kg * 9.8 m/s^2 F= 646.8N confidence assessment: 1
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21:16:45 ** ** The normal force is the force between and perpendicular to the two surfaces in contact. If the person was standing on the floor in equilibrium, the normal force would be equal and opposite to the person's 650 N weight. However during the jump the person is not in equilibrium. While in contact with the floor, the person is accelerating upward, and this implies a net upward force. This net force is comprised of the gravitational and normal forces: F_net = weight + normal force. Choosing the upward direction as positive, the person's weight is -650 N so F_net = -650 N + normal force. A quick solution: This net force must do enough work to raise the person's weight 1 meter while acting through only a .2 meter displacement, and must therefore be 5 times the person's weight. The normal force is equal to the net force plus the person's weight, so is 6 times the person's weight. The detailed reasoning is as follows: At the top of the jump the mass is 1 meter higher than at the bottom, so gravitational PE has increased by `dPE = 650 N * 1 meter = 650 Joules. The PE increase is due to the work done by the net force during the .2 meter interval before leaving the floor. Thus Fnet * (.20 meters) = PE increase. Since F_net = F_normal - 650 N we have ( F_normal - 650 N ) * (.20 m) = PE increase, and since PE increase is 650 Joules we have ( F_normal - 650 N ) * (.20 m) = 650 Joules. So (Fnormal - 650 N) * .2 meters = 650 Joules Fnormal - 650 N = 650 J / (.2 m) Fnormal = 650 J / (.2 m) + 650 N = 3250 N + 650 N = 3900 N. An average force of 3900 N is required to make this jump from the given crouch. The information given in this problem probably doesn't correspond with reality. A 3900 N force is equivalent to the force exerted by a 250-lb weightlifter doing a 'squat' exercise with about 600 pounds on his shoulders. It is extremely unlikely that anyone could exert this much force in a jump. More likely the 'crouch' required for a 1-meter jump would be significantly more than 20 cm. A 20-cm crouch is only about 8 inches and vertical jumps typically involve considerably more crouch than this. With a 40-cm crouch such a jump would require only half this total force and is probably feasible. **
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RESPONSE --> the energy is PE because it is not actually in motion and acting against force until the person jumps?
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