#$&* course PHY 121 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a The acceleration of the mass is a = F_net / m, so the velocity of the object changes by amount `dv = a * `dt = F_net / m * `dt. Since the initial velocity is zero, this will also be the final velocity: vf = F_net / m * `dt. From this and the fact that acceleration is constant (const. net force on const. mass implies const. acceleration), we conclude that vAve = (v0 + vf) / 2 = (0 + (F_net / m) * `dt) / 2 = F_net * `dt / (2 m). Multiplying this by the time interval `dt we have `ds = vAve `dt = (F_net * `dt) / (2 m) * `dt = F_net `dt^2 / (2 m). If we multiply this by F_net we obtain F_net * `ds = F_net * F_net * `dt^2 / (2 m) = F_net^2 * `dt^2 / (2 m). From our earlier result vf = F_net / m * `dt we see that KE_f = 1/2 m vf^2 = 1/2 m ( F_net / m * `dt)^2 = F_net^2 * `dt^2 / (2 m). Our final KE, when starting from rest, is therefore equal to the product F_net * `ds. Since we started from rest, the final KE of the mass on this interval is equal to the change in KE on the interval. We call F_net * `ds the work done by the net force. Our result therefore confirms the work-kinetic energy theorem: `dW_net = `dKE. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This has all just jumbled together in my head. I am not sure what is what. Im just not clear on which is which.
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Given Solution: `a** The system does positive work at the expense of its kinetic and/or potential energy. The work done by the system against all forces is `dW_net_BY. `dW_net_BY is equal and opposite to `dW_net_ON, which is in turn equal to `dKE, the change in the kinetic energy of the system. We conclude that `dW_net_BY = - `dKE. The change in KE is equal and opposite to the work done by the system against the net force acting on it. To consider the role of PE, we first review our formulation in terms of the work done ON the system: `dW_net_ON = `dKE. The work `dW_net_ON is the sum of the work done on the system by conservative and nonconservative forces: `dW_net_ON = `dW_cons_ON + `dW_NC_ON and `dW_cons_ON is equal and opposite to `dPE, the change in the system's PE. Thus `dW_net_ON = `dW_NC_ON - `dPE so that `dW_net_ON = `dW_cons_ON + `dW_NC_ON becomes `dW_NC ON - `dPE = `dKE so that `dW_NC_ON = `dPE + `dKE. Since `dW_NC_BY = - `dW_NC_ON, we see that -`dW_NC_BY = `dPE + `dKE so that `dW_NC_BY + `dPE + `dKE = 0. Intuitively, if the system does positive work against nonconservative forces, `dPE + `dKE must be negative, so the total mechanical energy PE + KE of the system decreases. (Similarly, if the system does negative work against nonconservative forces that means nonconservative forces are doing positive work on it, and its total mechanical will increase). As usual, you should think back to the most basic examples in order to understand all these confusing symbols and subscripts (e.g., if I lift a mass, which you know intuitively increases its gravitational potential energy, I do positive work ON the system consisting of the mass, the conservative force of gravity acts in the direction opposite motion thereby doing negative work ON the system, and the work done BY the system against gravity (being equal and opposite to the work done ON the system by gravity) is therefore positive). The equation -`dW_NC_BY = `dPE + `dKE isolates the work done by the system against nonconservative forces from the work it does against conservative forces, the latter giving rise to the term `dPE. If the system does positive work against conservative forces (e.g., gravity), then its PE increases. If the system does positive work against nonconservative forces (e.g., friction) then `dPE + `dKE is negative: PE might increase or decrease, KE might increase or decrease, but in any even the total PE + KE must decrease. The work done against a nonconservative force is done at the expense of at least one, and maybe both, the PE and KE of the system. (In terms of gravitational forces, the system gets lower or slows down, and maybe both, in order to do the work). If nonconservative forces do positive work on the system, then the system does negative work against those forces, and `dW_NC_ON is negative. Thus -`dW_NC_ON is positive, and `dPE + `dKE is positive. Positive work done on the system increases one or both the PE and the KE, with a net increase in the total of the two. (In terms of gravitational forces, the work done on system causes it to get higher or speed up, and maybe both.) 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: that is one example; another might be work to compress a spring STUDENT QUESTION ok, alot to absorb but I think I am getting there. So KE is equal to work done ON the system not BY the system...this is a little confusing. I was thinking that when we calculate the dw= fnet * 'ds we were caculated the work done BY the system not on the system.? INSTRUCTOR RESPONSE To be very specific `dW_net_ON = F_net_ON * `ds is the work done by the net force acting ON the system, where F_net_ON is the net force acting on the system. The work-kinetic energy theorem states that `dW_net_ON = `dKE If positive work is done on a system, it speeds up. If negative work is done on the system, it slows down. From the point of view of the system, if positive work is done by the system then the system has to 'use up' some of its kinetic energy to do the work, so it slows. Positive work done BY the system constitutes negative work being done ON the system. If part of the net force is conservative, then `dW_net_ON can be split into `dW_net_ON_cons and `dW_net_ON_noncons. The quantity `dPE, the change in PE, is defined to be equal and opposite to `dW_net_ON_cons. That is, `dPE = - `dW_net_ON_cons. It follows that `dW_net_ON = `dW_net_ON_noncons - `dPE, so that the work-kinetic energy theorem can be rewritten as `dW_net_ON_noncons - `dPE = `dKE. This is commonly rearranged to the form `dW_net_ON_NC = `dKE + `dPE. STUDENT COMMENT (confused by too many symbols) Once again this makes no since to me. All the symbols lost me
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Again, work done on and work done by just really confuse me. I cant tell the difference. ------------------------------------------------ Self-critique rating:
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Given Solution: `a** Bottom line: The system accelerates from zero to max KE then back to zero, defining an interval for which `dKE is positive and an interval for which `dKE is negative. The system starts and ends at rest so the total `dKE, from the beginning of the first interval to the end of the second, is zero. F_net_ave * `ds between the initial state of rest and max KE must therefore be equal and opposite to F_net_ave * `ds between max KE and the final state of rest. During the second interval the net force is the frictional force, which is assumed constant, i.e., the same no matter how far the rubber band was pulled back. During the second interval, therefore, F_net_ave remains constant, so it is the coasting displacement that varies with pullback. The coasting displacement is therefore proportional to the F * `ds total for work done by the rubber band on the system. More details: The F_`ds total for the rubber band is the work done to accelerate the rail to its maximum velocity v_max. Let's denote this simply by F_ave * `ds, where F_ave is understood to be the average force exerted by the rubber band (the rubber band force is at its maximum when the rubber band is pulled back, and decreases to 0 as it 'snaps back', accelerating the rail; so it makes sense to talk about the average rubber band force) and `ds is the displacement through which this force acts (i.e., the displacement from release until the rubber band loses contact with the rail). While in contact with the 'rail' the rubber band exerts its force in the direction of the system's motion and therefore does positive work. So F_ave * `ds is positive. The 'rail' then coasts to rest subject to the force of friction, which acts in the direction opposite motion and therefore does negative work. Assuming the frictional force f_frict to be constant, and using `ds_coast for the coasting displacement, the work done against friction is f_frict * `ds_coast. For simplicity of notation we will neglect the presence of the frictional force during the first interval, while the rubber band is in contact with the 'rail'. It isn't completely accurate to do so, but if the displacement during this interval is small compared to the coasting distance the error is small. A comment at the end will indicate how to easily modify these results. We will also neglect any other forces that might be acting on the system, so that the net force for the first phase is just the rubber band force, and for the second phase the net force is just the frictional force. Now, during the first interval the rail's KE changes from 0 to 1/2 m v_max^2, where m is its mass, so by the work-KE theorem F_ave * `ds = `dKE = 1/2 m v_max^2. During the second interval the rail's KE changes from 1/2 m v_max^2 to 0, so that f_frict * `ds_coast = -1/2 m v_max^2. Thus F_ave * `ds = - f_frict * `ds_coast so that the coasting displacement is `ds_coast = - (F_ave * `ds) / f_frict = (- 1 / f_frict) * F_ave * `ds. F_ave and f_friction are in opposite directions, so if F_ave is positive f_frict is negative, making -1 / f_frict negative and `ds_coast = (-1 / f_frict) * (F_ave * `ds) indicates a direct proportionality between `ds_coast and F_ave * `ds. The above relationship tells us that the coasting displacement is proportional to the F * `ds total for the force exerted by the rubber band. To correct the oversimplification of the given solution, if that oversimplification bothers you, you may proceed as follows (however if you find you don't completely understand the preceding you shouldn't confuse yourself with this until you do): To account for the frictional force while the rubber band is in contact with the rail, assuming that the frictional force is also present during the first phase, we can simply replace `ds_coast with `ds_coast + `ds. The f_frict * (`ds_coast + `ds) will be the actual quantity that is proportional to F_ave * `ds for the rubber band. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Again, all of this just jumbles in my head. This stuff Im just not getting at all. Its not clicking. Hopefully it will soon. ------------------------------------------------ Self-critique rating:2 ********************************************* Question: `qgen phy A person of mass 66 kg crouches then jumps to a height of .8 meters. From the crouched 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** STUDENT QUESTION I was not sure how to find the average force so I just multiplied the mass by 9.8. After looking at the solution I am still confused on finding the normal force of the object. INSTRUCTOR RESPONSE Imagine you are lying on the floor, with 2' x 2' piece of plywood resting on your chest, and a child about half your weight standing on the plywood. If the child does a vertical leap, what will that feel like to you? You can imagine the same with a person of your own size. If it's an average person with, say, a 20-inch vertical leap, what would that feel like? If it's someone with a 40-inch vertical leap, what would that feel like? In each case you are experiencing the normal force. When the person is just standing on the board . The normal force is just equal and opposite the person's weight and it's not difficult for you to manage. When the person jumps, his or her legs push down, hard, and the board (and you by extension) have to push back. In the case of the 40-inch leap, you'll probably find that to be at least uncomfortable. For the person to leap the floor has to push up hard enough to result in a net force equal to mass * acceleration (F_net = m a). Gravity is pulling down, so the net force is (net force) = (floor push) - (weight), so (floor push) = (net force) + (weight). STUDENT QUESTION this has .2 I am not sure were .2 came from, but in this problem we have .8 INSTRUCTOR RESPONSE The person's altitude changes by `dy = 1 meter, from the .2 meter crouch to the .8 meter height. So PE increases by `dPE = weight * `dy. In this case `dPE is about 650 Joules. The person's feet stay in contact with the floor for the first .2 meters. Only during this interval is force being exerted between feet and floor. So the work required to raise the PE is done during the .2 meter displacement. Thus F_net * .2 meters = 650 Joules and F_net = 650 J / (.2 m) = 3250 N, approx.. F_net = F_normal + weight, so F_normal, the force between feet and floor, is F_net + weight. This comes out around 3900 N. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `quniv phy text prob 4.46 (11th edition 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: