course Phy 121 ddɞXfassignment #014
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18:23:31 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 --> By multiplying the accerleration by the time interval, you obtain the change in velocity. The object travels : vAve * `dt = `ds When you multiply the net force by the distance traveled you get .5mv^2 or the KInetic Energy. .5mv^2
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18:24:21 **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 --> ok
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18:26:01 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 --> The change in the KE will always be equal to the opposite of the potential energy plus the nonconservative forces.
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18:26:54 ** 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 --> ok
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18:30:33 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 --> The work done to stretch the rubber band and the work done on the rail by the band is not represented as a linear relationship. The PE of the stretched band is converted to the KE of the rail. Frictional forces decrease the KE and cannot be reclaimed. W = .5kx^2
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18:31:06 ** 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 --> ok
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18:31:35 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 PE + KE = 0
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18:33:22 ** 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 --> Change in v^2 under influnece of the rubber band is equal and opposite to the change in v^2 as it works against friction.
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18:33:25 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 -->
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18:33:27 ** the normal force is the force between and perpendicular to the two surfaces in contact, which would be 646.8N if the jumper was in equilibrium. However during the jump this is not the case, and the normal force must be part of a net force that accelerates the jumper upward. In a nutshell the 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 person still has to support his weight so the normal force must be 6 times the person's weight. The detailed reasoning is as follows: To solve this problem you have to see that the average net force on the jumper while moving through the `dy = 20 cm vertical displacement is equal to the sum of the (upward) average normal force and the (downward) gravitational force: Fnet = Fnormal - m g. This net force does work sufficient to increase the jumper's potential energy as he or she rises 1 meter (from the .20 m crouch to the .8 m height). So Fnet * `dy = PE increase, giving us ( Fnormal - m g ) * `dy = PE increase. PE increase is 66 kg * 9.8 m/s^2 * 1 meter = 650 Joules approx. m g = 66 kg * 9.8 m/s^2 = 650 Newtons, approx.. As noted before `dy = 20 cm = .2 meters. 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. This 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 without the additional weight. A 20-cm crouch is only about 8 inches and vertical jumps are typically done with 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 -->
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course Phy 121 wINNċxѣStudent Name:
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19:09:04 `q001. Note that this assignment contains 5 questions. . Suppose that a net force of 10 Newtons acts on a 2 kg mass for 3 seconds. By how much will the velocity of the mass change during these three seconds?
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RESPONSE --> Fnet / m = a a * `dt = `dv 10N / 2kg = 5m/s^2 5m/s^2 * 3 sec = 15m/s
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19:09:11 The acceleration of the object will be accel = net force / mass = 10 Newtons / (2 kg) = 5 m/s^2. In 3 seconds this implies a change of velocity `dv = 5 m/s^2 * 3 s = 15 meters/second.
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RESPONSE --> ok
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19:12:38 `q002. By how much did the quantity m * v change during these three seconds? What is the product Fnet * `dt of the net force and the time interval during which it acted? How do these two quantities compare?
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RESPONSE --> 2kg * 15m/s = 30 kgm/s 10N * 3second = 30N sec The quantities are the same but I'm not understanding the units.
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19:13:00 Since m remained constant at 2 kg and v changed by `dv = 15 meters/second, it follows that m * v changed by 2 kg * 15 meters/second = 30 kg meters/second. Fnet *`dt is 10 Newtons * 3 seconds = 30 Newton * seconds = 30 kg meters/second^2 * seconds = 30 kg meters/second. The two quantities m * `dv and Fnet * `dt are identical.
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RESPONSE --> ok
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19:17:24 `q003. The quantity m * v is called the momentum of the object. The quantity Fnet * `dt is called the impulse of the net force. The Impulse-Momentum Theorem states that the change in the momentum of an object during a time interval `dt must be equal to the impulse of the average net force during that time interval. Note that it is possible for an impulse to be delivered to a changing mass, so that the change in momentum is not always simply m * `dv; however in non-calculus-based physics courses the effective changing mass will not be considered. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be the impulse of the force?
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RESPONSE --> 3000m/s
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19:18:18 The impulse of the force will be Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. Note that the 1200 kg mass has nothing to do with the magnitude of the impulse.STUDENT COMMENT: That's a little confusing. Would it work to take the answer I got of 3234 N and add back in the weight of the person at 647 N to get 3881? INSTRUCTOR RESPONSE: Not a good idea, though it works in this case. Net force = mass * acceleration. That's where you need to start with problems of this nature.Then write an expression for the net force, which will typically include but not be limited to the force you are looking for. *&*&
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RESPONSE --> 3000 kg meters/sec ok
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19:22:07 `q004. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be change in the velocity of the vehicle?
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RESPONSE --> First find a 2000N / 1200kg = 1.67m/s^2 1.67 m/s^2 * 1.5 sec = 2.5 m/s
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19:22:51 The impulse of the 2000 Newton force is equal to the change in the momentum of the vehicle. The impulse is impulse = Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. The change in momentum is m * `dv = 1200 kg * `dv. Thus 1200 kg * `dv = 3000 kg m/s, so `dv = 3000 kg m/s / (1200 kg) = 2.5 m/s. In symbols we have Fnet * `dt = m `dv so that `dv = Fnet * `dt / m.
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RESPONSE --> ok
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19:25:22 `q005. Use the Impulse-Momentum Theorem to determine the average force required to change the velocity of a 1600 kg vehicle from 20 m/s to 25 m/s in 2 seconds.
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RESPONSE --> `dv = 20m/s 20m/s * 1600kg = 32000 kg m/s 3200kg m/s / 2 second = 16000 Newtons
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19:26:23 The vehicle changes velocity by 5 meters/second so the change in its momentum is m * `dv = 1600 kg * 5 meters/second = 8000 kg meters/second. This change in momentum is equal to the impulse Fnet * `dt, so Fnet * 2 sec = 8000 kg meters/second and so {} Fnet = 8000 kg meters/second / (2 seconds) = 4000 kg meters/second^2 = 4000 Newtons. In symbols we have Fnet * `dt = m * `dv so that Fnet = m * `dv / `dt = 1600 kg * 5 m/s / ( 2 s) = 4000 kg m/s^2 = 4000 N.
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RESPONSE --> Oh, for some reason I was thinking the change was 20 not 5 m/s.I think I still get this.
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