course Phy 231 ????Q???+???p???assignment #014014. `query 14
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17:41:55 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 --> Acceleration multiplied by the time interval yields the change in velocity. Knowing the initial velocity and the change in velocity will yield the final velocity. With final velocity, initial velocity and acceleration, the displacement can be found. vf = `dv + v0 `ds = (v^2-v0^2)/(2*a) Multiplying the net force by the distance travels is the work done on the object. This work is equal to the gain in kinetic energy of the object. confidence assessment: 3
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17:43:42 **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 --> self critique assessment: 3
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17:48:38 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 --> When a system does work against non-conservative forces the energy that is expended in KE is not converted to PE which can be recovered. When a system does work against a conservative force, the energy that is expended in KE is converted to PE and can at least be partially recovered. confidence assessment: 3
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17:50:55 ** 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 --> I should have said that in a conservative situation that all KE converts to PE and that the PE can be fully recovered to KE self critique assessment: 2
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18:34:14 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 converts KE into PE which is then transformed into thermal energy and Kiinetic energy which is applied to the rail. The work done by the rail against friction and thus from the rubber band on the rail far exceeds the force of friction. The force of friction against the rail is constant and does contribute to bringing the rail to a stop. confidence assessment: 2
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18:37:09 ** 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 --> Right. The force of the rubber band acting on the rail is equal to the work done by the rail against friction through the motion of the rail except for some loss to thermal energy. self critique assessment: 2
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18:45:14 Why should the distance traveled by the rail be proportional to the F * `ds total for the rubber band?
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RESPONSE --> The distance traveled by the rail is proportional to the F*`ds of the rubber band because ]a graph of pullback vs. observed distances that the rail traveled revealed a nearly linear graph. This implies proportionality between F and x, which would also mean a proportionality in x and F*`ds. confidence assessment: 3
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18:50:36 ** 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 --> I was on the right track, but I didn't quite reach the sophistication that I needed here. The important thing to remember here is tha F`ds is proportional to the change in v^2 and that and that f`ds is also proportional to the change in vf^2. self critique assessment: 2
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18:50:52 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 --> confidence assessment: 3
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18:51:05 ** 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 --> self critique assessment: 3
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18:51:16
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RESPONSE --> self critique assessment: 3
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21:04:53 univ phy text prob 4.42 (4.40 in 10th edition) Mercury lander near surface upward thrust 25 kN slows at rate 1.2 m/s^2; 10 kN speeds up at .8 m/s^2; what is weight at surface?
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RESPONSE --> In the case of the 25kN thrust upward and the slowing down at 1.20m/s^2, the net force is in the upward direction. Fnet = 25kN - w and Fnet = m * 1.2m/s^2 so: 25kN - w = m * 1.2m/s^2 Fnet = 10kN -w and Fnet = m * -.8m/s^2 so: 10kN - w = m * -.8m/s^2 Solving each of the combined equations for -w and setting them equal to each other: m * 1.2m/s^2 - 25kN = m* -.8m/s^2 -10kN m * 1.2m/s^2 = m * -.8m/s^2 + 15kN 2m/s^2 * m = 15000N m = 15000N/2m/s^2 = 7500kg In the first case: Fnet = 7500kg * 1.2m/s^2 Fnet = 9000N 25000N - w = 9000N, so w = 16000N Verifying In the second case: Fnet = 7500kg * -.8m/s^2 Fnet = -6000N 10000N - w = -6000N w = 16000N So the weight of the lander near the surface is 16000N confidence assessment: 3
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21:05:29 ** If a landing craft slows then its acceleration is in the direction opposite to its motion, therefore upward. If it speeds up while landing that its acceleration is in the direction of its motion, therefore downward. If the upward motion is taken as the positive direction, then the acceleration under a thrust of 25 kN is + 1.2 m/s^2, and the acceleration when under thrust of 10 kN is - .8 m/s^2. In either case m * a = net force. Net force is thrust force + gravitational force. 1 st case, net force is 25 kN so m * 1.2 m/s/s + m * g = 25 kN. 1 st case, net force is 10 kN so m * (-.8 m/s/s ) + m * g = 10 kN. Solve these equations simultaneously to get the weight m * g (multiply 1 st eqn by 2 and 2d by 3 and add equations to eliminate the first term on the left-hand side of each equation; solve for m * g). The solution is m * g = 16,000 kN. Another solution: In both cases F / a = m so if upward is positive and weight is wt we have (25 kN - wt) / (1.2 m/s^2) = m and (10 kN - wt) / (-.8 m/s^2) = m so (25 kN - wt) / (1.2 m/s^2) = (10 kN - wt) / (-.8 m/s^2). Solving for wt we get 16 kN. **
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RESPONSE --> self critique assessment: 3
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