course Phy 231 6/24 5 014. `query 14*********************************************
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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): OK Self-critique rating #$&* OK ********************************************* Question: `q Define the relationship between the work done by a system against nonconservative forces, the work done against conservative forces and the change in the KE of the system. How does PE come into this relationship? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Work by nonconservative forces on the system + work by conservative forces on the system = net work on the system = change in KE Change in KE + change in PE = work by nonconservative forces on the system Confidence rating #$&*: 3
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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.) 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `qclass notes: rubber band and rail How does the work done to stretch the rubber band compare to the work done by the rubber band on the rail, and how does the latter compare to the work done by the rail against friction from release of the rubber band to the rail coming to rest? Your solution: If the situation were ideal with no other forces acting on the system, the work done to stretch the rubber band would be equal to the work done by the rubber band on the rail. This would also be equal to the work done by the rail against friction. Confidence rating #$&*: 3
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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): This works because the forces are equal and opposite, right??????
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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): OK Self-critique rating #$&* OK ********************************************* Question: `qgen phy A person of mass 66 kg crouches then jumps to a height of .8 meters. From the crouched position to the point where the person leaves the ground the distance is 20 cm. What average force is exerted over this 20-cm distance? Your solution: Normal force + weight = net force Weight = 66kg(-9.8m/s^2) = -646.8N Change in gravitational PE = 646.8N *1m = 646.8J Ds for PE is 20 cm or .2m PE increase = Fnet*ds 646.8J = Fnet*.2m and Fnet = normal -646.8N 646.8J = normal - 646.8N * .2m Normal – 646.8N = 3234N Normal = 3880.8N confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 minus 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `quniv phy text prob 4.46 (11th edition 4.42) (4.40 in 10th edition) Mercury lander near surface upward thrust 25 kN slows at rate 1.2 m/s^2; 10 kN speeds up at .8 m/s^2; what is weight at surface? Your solution: Upward thrust: -25000N Acceleration = -1.2m/s^2 10000N; .8m/s^2 Net force = thrust+gravity 25000 = - 1.2m/s^2 (m) +g(m) 10000 = .8m/s^2(m) +g(m) solve the system of equations 5M*g = 50000N+30000N 5mg = 80000N Mg = 16000N or 16kN Confidence rating #$&*: 3
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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 = 16,000 N. 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): OK Self-critique rating #$&* OK "