#$&* course Phy 231 2/25 10:50pm 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): I think my answer corresponds pretty well with your explanation. I skipped over the definition of `dKE as 1/2 m*vf^2 - m*v0^2, but I see why it's important to keep the meaning in mind. ------------------------------------------------ Self-critique rating: ********************************************* 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: The sum of all work done BY a system is equal to `dKE. This is equal and opposite to the sum of all work done ON a system. The work done by a system against nonconservative forces is lost in KE and not recovered in PE. The work done against conservative forces is lost in KE but recovered in PE. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK, so when I said that the work done on friction must exceed the work done by the rubber band on the rail, that's not really true. This goes back to a previous assignment where you said friction behaves as though it COULD do a certain amount of work, but in reality it can't do more work on the object than the object does on it. I guess I was thinking friction must overpower the work of the rubber band. And I'm still thinking that mathematically, that is so, but in reality, it does exactly as much work as it needs to do to stop the rail, and then has no reason to do more, even if it is capable of doing more. ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhy should the distance traveled by the rail be proportional to the F * `ds total for the rubber band? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Ideally, the work required to stretch the rubber band (F on the rubber band * `ds the rubber band stretches) becomes the work the rubber band does on the rail (F on the rail * `ds the rail travels), and in turn, friction does equal and opposite work on the rail until it stops. Since the work for each portion of this process is equal to a force times a displacement, the displacement in each case is proportional to the work done (assuming a constant force.) And the work is equal in magnitude for each portion, so that means the displacement fo the rail should also be proportional to the initial work done. confidence 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. SYNOPSIS The 'rail' is brought to rest by a pretty-much-constant frictional force. The work it does against this force is therefore proportional to how far it slides. The 'rail' comes to rest when the frictional force has dissipated its initial kinetic energy, which was obtained from the rubber band. So initial KE, which is pretty much equal to F * `ds for the rubber band, is proportional to how far the 'rail' slides. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I'm a bit mixed up in how I'm thinking about the ""portions"" (intervals) of this process. I was thinking of (1) stretching the rubber band (2) the rubber band acting on the rail and (3) friction acting on the rail. But you divide it into just two intervals, and that confuses me. Is the work required to stretch the band and the work of the rubber band on the rail the same thing? You say the work of the band on the rail is what causes it to accelerate to max, and that makes sense, but isn't friction working during that time too? (Or wait, you address this-- so max velocity is at the exact moment of release? It is accelerating WHILE the rubber band is acting on it due to the force of the stretch, and we're ignoring friction during that time. But after release, that force ceases to act, and the only force and work on the rail is friction from that point on. Right?) ------------------------------------------------ Self-critique rating: Question (Openstax) : `prin, `gen: Consider the baby being weighed in Figure 4.34. (a) What is the mass of the child and basket if a scale reading of 55 N is observed? (b) What is the tension T1 in the cord attaching the baby to the scale? (c) What is the tension T2 in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg? (d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible. Baby supported from scale, scale in turn supported from ceiling. Figure 4.34 A baby is weighed using a spring scale. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The weight of the baby is the force exerted on it by gravity. That force is the product m * g of the baby's mass and the acceleration of gravity. Thus m * g = 55 Newtons so that m = 55 Newtons / g = 55 kg m/s^2 / (9.8 m/s^2) = 5.6 kilograms. The baby is not accelerating so the net force on it is zero. The only forces acting on it are the 55 Newton gravitational force, which acts downward, and the tension in the cord, which acts upward. Your figure should show a downward force of 55 Newtons and the upward force T_1. Choosing upward as the positive direction, then, we have T_1 - 55 Newtons = 0 so that T_1 = 55 Newtons. The net force on the scale is also zero, since it is not accelerating. The tension in cord 1 acts downward on the scale. So the forces on the scale are -T_1, T_2 and the gravitational force -0.50 kg * 9.8 m/s^2 = -4.9 Newtons. Your figure should show the downward force -T_1, the downward for -4.9 N and the upward force T_2. Thus we have -T_1 + T_2 + -4.9 Newtons = 0 so that T_2 = T_1 + 4.9 Newtons = 55 Newtons + 4.9 Newtons = 59.5 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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** Bascially the person rises 1 meter from the crouched position. This requires a change in gravitational PE, which is easily calculated. This change requires that work be done, and work can only be done while the player's feet are on the floor. Afterwards they are not in contact with anything and no upward force can be exerted. The upward force is exerted by the floor in reaction to downward force exerted by the player's legs. The upward force is exerted while the player moves 20 cm. Having calculated the required PE change you know how much work must be done during the 20 cm interval. So you know the displacement and need only find the average force. More formally, for the interval from the floor to the highest point of the jump: `dW_noncons_ON = F_floor * 0.2 m `dPE = 1 m * 66 kg * 9.8 m/s^2 `dKE = 0 (for the interval from crouched position to highest point the player is at rest at the initial and the final instant). `dW_nc_ON = `dPE + `dKE so F_floor * 0.2 m = 1 m * 66 kg * 9.8 m/s^2 + 0 F_floor = 1 m * 66 kg * 9.8 m/s^2 / 0.2 m = 3200 N, approx.. 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 of another equivalent explanation is as follows: `dW_nc_ON = `dPE + `dKE. `dKE is zero, since the person is at rest before the jump and at the top of the jump. So the work done by the nonconservative force exerted by the floor must be equal to `dPE. 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. Thus `dPE = +650 J. The PE increase is due to the work done by the normal force (exerted by the floor) during the .2 meter interval before leaving the floor. Thus F_normal * (.20 meters) = PE increase. so that F_normal = PE increase / (.20 m) = 650 J / (.20 m) = 3300 N. An average force of 3300 N is required to make this jump from the given crouch. The information given in this problem probably doesn't correspond with reality. A 3300 N force is equivalent to the force exerted by a 250-lb weightlifter doing a 'squat' exercise with about 500 pounds on his shoulders. It is unlikely that a 66 kg player would exert this much aveage force throughout a jump (the force during the jump will be variable, and the maximum force might well exceed 3300 N; however as the player accelerates the force will likely decrease). 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 force calculated here, 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 exerted by the person on the board, which in turn exerts this force (plus a little bit more, due to the weight of the board itself) on you. You in turn exert an equal and opposite upward force, which is what causes the person to rise. When the person is just standing on the board, the normal force is just equal and opposite the person's weight (we will regard the weight of the board as negligible) 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 starts from a .2 meter crouch. The player'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 player'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 increase 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. SYNOPSIS It takes about 650 Joules to raise a 650 N person to a height of 1 meter. The person cannot exert any lifting force after leaving the floor, which happens after traveling .20 meters. So the person has to do 650 Joules of work while traveling .20 meters. This requires 3300 N of force. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Question (Openstax): `gen Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration due to gravity. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The acceleration of gravity is 9.8 m/s^2, so four times this acceleration is 4 * 9.8 m/s^2 = 39.2 m/s^2. To produce this acceleration the net force on the jumper must be equal to m * a, where m is the 70 kg mass and a is the required 39.2 m/s^2 acceleration. The forces on the jumper are the force exerted by gravity (i.e., the jumper's weight) and the upward force exerted by the floor. To produce the required upward force the jumper must exert an equal and opposite downward force on the floor. If F_floor is the force exerted by the floor, then the net force on the jumper is F_net = F_floor - m * g where we have chosen the upward direction as positive. It follows that F_floor = F_net + m * g = 70 kg * 39.2 m/s^2 + 70 kg * 9.8 m/s^2 = 3470 Newtons. STUDENT QUESTION According to the equation Fnet=m*a (70kg*39.2m/s^2 =2744N) it takes 2744N to accelerate a mass of 70 kg to 39.2m/s^2. So why do we have to add the weight of the jumper as well? I thought that was included in the original equation? INSTRUCTOR RESPONSE The net force is equal to the mass * acceleration. The net force is equal to the sum of the forces acting on the jumper. The normal force and the force of gravity both act on the jumper. So net force = mass * acceleration = 70 kg * 4 m/s^2 = 2800 N. net force = normal force + gravitational force Gravitational force is -70 kg * 9.8 m/s^2 = -700 N, approx.. Thus normal force = net force - gravitational force = 2800 N - (-700 N) = 3500 N, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: The force of Mercury's gravity on the lander is acting in the positive direction, down toward the surface. (a) Fnet=Fgrav+Fthrust m*a=Fgrav-25kN m*(a-1.2m/s^2)=Fgrav-25kN
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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. ** ANOTHER NEARLY IDENTICAL SOLUTION, WHICH ALSO SOLVES FOR THE ACCELERATION OF GRAVITY ON MERCURY F_net = m a. Choose upward as positive. At 25 000 N we have F_net = m * 1.2 m/s^2, and at 10 000 N we have F_net = m * (-.8 m/s^2). In the first case F_net = 25 000 N - m g_planet and in the second F_net = 10 000 N - m g_planet. So 25 000 N - m g_planet = m * 1.2 m/s^2 and 10 000 N - m g_planet = m * (-.8 m/s^2). We can eliminate the unknown g_planet by subtracting the second equation from the first, to get 15 000 N = m * 2 m/s^2 so that m = 15 000 N / (2 m/s^2) = 7500 kg. Then the first equation becomes 25 000 N = 7500 kg ( 1.2 m/s^2 + g_planet) so that 7500 kg * g_planet = 16 000 N and g_planet = 2.13 m/s^2. So the weight of the lander is 7500 N * 2.13 m/s^2 = 16 000 N. The net force with 25 000 N thrust is 9 000, resulting in acceleration 9 000 N / (7500 kg) = 1.2 m/s^2. With 10 000 N thrust the resulting acceleration is -6000 N / (7500 kg) = -.8 m/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): So 25 kN and 10kN represent NET force on the craft for those instances, not the force of the the landing equipment to speed or slow the descent (ie to apply additional force to the net force?)