#$&* course phy201 014. Potential energy; conservative and non-conservative forces. *********************************************
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Given Solution: The weight of the automobile is 1500 kg * 9.8 m/s^2 = 14,700 Newtons. The weight component parallel to the incline is therefore very close to the small-slope approximation weight * slope = 14700 Newtons * .03 = 441 Newtons (small-slope approximation). If no other forces act parallel to the incline then the net force will be just before 441 Newtons and the work done by the net force will be `dWnet = 441 Newtons * 200 meters = 88200 Joules. [ Note that this work was done by a component of the gravitational force, and that it is the work done on the automobile by gravity. ] The net work on a system is equal to its change in KE. Since the automobile started from rest, the final KE will equal the change in KE and will therefore be 88200 Joules, and the final velocity is found from 1/2 m vf^2 = KEf to be vf = +_`sqrt(2 * KEf / m) = +-`sqrt(2 * 88200 Joules / (1500 kg) ) = +-`sqrt(2 * 88200 kg m^2/s^2 / (1500 kg) ) = +- 10.9 m/s (approx.). Since the displacement down the ramp is regarded as positive and the automotive will end up with a velocity in this direction, we choose the +10.9 m/s alternative. STUDENT QUESTION: I assume the perpendicular force is balanced beause of the normal force downward of gravity and mass and then upward from the road. INSTRUCTOR RESPONSE: Good, but there's a little more to it: The normal force balances the component of the gravitational force which is perpendicular to the road. The component of the gravitational force parallel to the incline is the for that tends to accelerated objects downhill. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I used the wrong formula to find the vf of the car, I should have done: Kef = .5* m * vf2 88200 J = .5 * 1500 kg * vf2 Vf = 10.84 m/s ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. If the automobile in the preceding problem is given an initial velocity of 10.9 m/s up the ramp, then if the only force acting in the direction of motion is the force of gravity down the incline, how much work must be done by the gravitational force in order to stop the automobile? How can this result be used, without invoking the equations of motion, to determine how far the automobile travels up the incline before stopping? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I have no idea, after looking at the solution: KE0 = .5 m v2 .5 * 1500 * 10.92 = 89107 J Gravitational force on car on incline = 441 N dWnet = -441N * 200 = -88200 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This is an application of the work-kinetic energy theorem. In words, this theorem says that the change in KE is equal to the work done by the net force acting ON the system In symbols, this is expressed `dW_net = `d(KE). KE is kinetic energy, equal to 1/2 m v^2. The automobile starts out with kinetic energy KEinit = 1/2 m v0^2 = 1/2 (1500 kg) ( 10.9 m/s)^2 = 88000 kg m^2/s^2 = 88000 Joules. The gravitational force component parallel to the incline is in this case opposite to the direction of motion so that gravity does negative work on the automobile. Since the change in KE is equal to the work done by the net force acting ON the system, if the gravitational force component parallel to the incline does negative work the KE of the object will decrease. This will continue until the object reaches zero KE. As found previously the gravitational force component along the incline has magnitude 441 Newtons. In this case the forces directed opposite to the direction motion, so if the direction up the incline is taken to be positive this force component must be -441 N. By the assumptions of the problem this is the net force exerted on the object. Acting through displacement `ds this force will therefore do work `dWnet = Fnet * `ds = -441 N * `ds. Since this force must reduce the KE from 88,000 Joules to 0, `dWnet must be -88,000 Joules. Thus `dW_net = -441 N * `ds = -88,000 Joules and `ds = -88,000 J / (-441 N) = 200 meters (approx.). Had the arithmetic been done precisely, using the precise final velocity found in the previous exercise instead of the 10.9 m/s approximation, we would have found that the displacement is exactly 200 meters. STUDENT QUESTION I set `dKE = Fnet * `ds and had `dKE as negative so I came up with the correct solution but just above you say that work is negative. I dont understand how work is negative, especially going by the equation because I thought work was opposite `dKE. INSTRUCTOR RESPONSE You're thinking about exactly the right things. The specific statement of the work-KE theorem is that the work done by the net force acting ON the system is equal to the change in the kinetic energy of the system. This is abbreviated `dW_ON_net = `dKE. Rather that talking about 'the work', it's very important to get into the habit of labeling 'the work' very specifically. You have two basic choices. You can think in terms of the work done on the system or object by a force the work done by the system or object against a force The two are equal and opposite. Note that the words 'on' and 'by' modify the word 'system', not the word 'force'. The key phrases are 'on the system' and 'by the system'. In the present case if you were to choose to think in terms of the work done by the net force exerted by object, then this force would be labeled `dW_BY_net and would be equal and opposite to `dW_ON_net, the work exerted by the net force acting on the object. Formally we would have `dW_BY_net = - `dW_ON_net so that `dW_BY_net = - `dKE. This last equation is often written `dW_BY_net + `dKE = 0, and is another equivalent formulation of the work-kinetic energy theorem. STUDENT COMMENT I am getting all the equations mixed up is there any way you can just send the different equations? I understand the 4 from the major quiz. I can do the algebra I just dont know which equation to plug it in for. INSTRUCTOR RESPONSE Physics is about more than figuring out what to plug into what equation. It's necessary to understand the words and the concepts to know which equation to plug into. In other words, the concepts are what keep us from getting the equations mixed up. However I have observed in your work that you do very well with the algebra. So the equations might well be your most appropriate starting point. You can use the equations to understand the words and the concepts, just as less algebraically adept students might use the words and concepts to understand the equations. The relevant relationships here are `dW_net = `dKE and KE = 1/2 m v^2. The relationship `dW_BY_net = - `dW_ON_net is also invoked in the additional comments at the end, which mention an alternative formulation of the work-kinetic energy theorem. However this relationship is not used in solving this particular problem. STUDENT COMMENT OK, I understand the solution and will use _on and _by descriptors in my answers from now on. INSTRUCTOR RESPONSE Good. Remember that ON and BY are adjectives applied to the word 'system', not to the word 'force'. That is, you have to determine whether the force is acting ON the system, or is exerted BY the system. Your choice of point of view will determine whether you use the equation `dW_NC_ON = `dPE + `dKE or `dW_NC_BY + `dPE + `dKE = 0. STUDENT COMMENT Ok. I see why my ds was negative. The F is negative in this system because it is working against the positive motion of the car UP the ramp. For every force, there is an equal and OPPOSITE force. INSTRUCTOR RESPONSE Good. If you assume the positive direction to be up the incline, F_net is negative, as you say, but the specific reason is slightly different than the one you give. It's good to think in terms of equal and opposite forces, but the motion of the car is not a force. In this case it really just comes down to signs: The force used to calculate `dW_net was the net force acting on the car. That force acts down the incline, in the direction opposite motion. Therefore F_net and `ds are of opposite sign, and the net force acting on the car does negative work. This decreases the KE, as your solution indicates. The question of whether `ds is positive or negative depends on which direction you choose for the positive direction. In your solution you apparently thought of upward as the positive direction; you should have explicitly stated this. Relative to this choice F_net is negative. As I mentioned, you should have declared the positive direction in your solution. You could have chosen either upward (which is the direction of the displacement, and is the direction you implicity chose) or downward (which is the direction of the net force) to be the positive direction. Either choice of positive direction would have been perfectly natural. If you had chosen 'down the incline' to be the positive direction, then `ds would have been negative (therefore opposite to the downward direction, so up the incline). In either case, `ds would have been up the incline. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 1 ********************************************* Question: `q003. If the automobile in the previous example rolls from its maximum displacement back to its original position, without the intervention of any forces in the direction of motion other than the parallel component of the gravitational force, how much of its original 88200 Joules of KE will it have when it again returns to this position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It will have all of its KE or kinetic energy confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In the first exercise in the present series of problems related to this ramp, we found that when the automobile coasts 200 meters down the incline its KE increases by an amount equal to the work done on it by the gravitational force, or by 88200 Joules. Thus the automobile will regain all of its 88200 Joules of kinetic energy. To summarize the situation here, if the automobile is given a kinetic energy of 88200 Joules at the bottom of the ramp then if it coasts up the ramp it will coast until gravity has done -88200 Joules of work on it, leaving it with 0 KE. Coasting back down the ramp, gravity works in the direction of motion and therefore does +88200 Joules of work on it, thereby increasing its KE back to its original 88200 Joules. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q004. Explain why, in the absence of friction or other forces other than the gravitational component parallel to incline, whenever an object is given a kinetic energy in the form of a velocity up the incline, and is then allowed to coast to its maximum displacement up the incline before coasting back down, that object will return to its original position with the same KE it previously had at this position. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The car initially had some KE. The gravitational component parallel to the incline is in the direction opposite to the direction of motion up the incline and therefore does negative work ON the object as it travels up the incline. The gravitational component is the net force on the object, so the work done by this net force on the system causes a negative change in KE, which eventually decreases the KE to zero so that the object stops for an instant. This happens at the position where the work done BY the net force is equal to the negative of the original KE. The gravitational component parallel to the incline immediately causes the object to begin accelerating down the incline, so that now the parallel gravitational component is in the same direction as the motion and does positive work ON the system. At any position on the incline, the negative work done by the gravitational component as the object traveled up the incline from that point, and the positive work done by this force as the object returns back down the incline, must be equal and opposite. This is because the displacement up the incline and the displacement down the incline are equal and opposite, while the parallel gravitational force component remains the same. Thus the Fnet * `ds products for the motion up and the motion down equal and opposite. When the object reaches its original point, the work that was done on it by the net force, as it rolled up the incline, must be equal and opposite to the work done on it while coasting down the incline. Since the work done on the object while coasting up the incline was the negative of the original KE, the work done while coasting down, being the negative of this quantity, must be equal to the original KE. Thus the KE must return to its original value. STUDENT QUESTION I still really dont understand how it can return back to its original position because of what we saw in class It never returned back to its original position. INSTRUCTOR RESPONSE There are a number of situations in which an object doesn't return to its original position. The one that's relevant to this situation: When you rolled the ball up the single incline, it slowed, came to rest for an instant, and then rolled back down. It did return to its initial position. Of course when it got there is was moving pretty fast so if you didn't stop it, it kept going until something else did. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. As the object travels up the incline, does gravity do positive or negative work on it? Answer the same question for the case when the object travels down the incline. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Gravity does negative work on the car as it travels up the incline the gravity is resisting the movement of the car up the incline. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: As the object travels up the incline the net force is directed opposite its direction of motion, so that Fnet_ON and `ds have opposite signs and as a result `dWnet = Fnet_ON * `ds must be negative. As the object travels down the incline the net force is in the direction of its motion so that Fnet_ON and `ds have identical signs and is a result `dWnet = Fnet_ON * `ds must be positive. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q006. If positive work is done on the object by gravity, will it increase or decrease kinetic energy of the object? Answer the same question if negative work is done on the object by gravity. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If positive work is done it will increase the kinetic energy of the car. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The KE change of an object must be equal to the work done ON the system by the net force. Therefore if positive work is done on an object by the net force its KE must increase, and if negative work is done by the net force the KE must decrease. STUDENT QUESTION (instructor comments in parentheses) Ok, after reading a couple of time I get it dw will be equal to KE. If dw is - Ke will be - . KE can't be negative; `dW is not equal to KE, but to `dKE, the change in KE. `dKE can certainly be positive or negative (or zero), depending on the situation. I am still a little unclear about if the dw done on an object is negative then what direction is it moving?? The sign of `dW by the net force does not determine the direction of motion of the object. It determines only the change in its kinetic energy. In the present case, the net force is the component of gravity along the incline. The direction of motion of the object determines whether this force is in the direction of motion or opposite that direction, and so determines whether the displacement is in the direction of motion (implying positive work) or opposite the direction of motion (implying negative work). The direction of motion thus determines, for this situation, whether `dW_net is positive or negative. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. While traveling up the incline, does the object do positive or negative work against gravity? Answer the same question for motion down the incline. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As the car travels up the incline it does positive work against gravity. The car is opposing gravity. As the car travels down the incline it does negative work against gravity because it is working with the gravitational force. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If the object ends up in the same position as it began, the work done on the object by gravity and work done by the object against gravity must be equal and opposite. Thus when the object does positive work against gravity, as when it travels up the incline, gravity is doing negative work against the object, which therefore tends to lose kinetic energy. When the object does negative work against gravity, as when traveling down the incline, gravity is doing positive work against the object, which therefore tends to gain kinetic energy. STUDENT COMMENT: A little shaky on this problem because I feel its easy to get confused on the positive and negative. INSTRUCTOR RESPONSE This is the most common point of confusion at this stage of the course. To sort out positive and negative, you would answer the following questions: Are you thinking about the work done ON the system or BY the system (i.e., are you thinking about the forces acting ON the system or a forces exerted BY the system)? The ON and the BY are equal and opposite. Whichever force you are thinking about, it does positive work when it is in the direction of motion and negative work when it is opposite the direction of motion. STUDENT QUESTION Ok, so i get his really mixed up. The work done BY the object is positve, against gravity which is doing negative work ON the object going up the inlcine. When going down the incline work done BY the object is negative as work done ON the object by gravity is positive. Is this right? INSTRUCTOR RESPONSE Your statement is correct. And, until it 'clicks', this is certainly confusing. It takes most students a few assignments before this becomes clear. You are progressing nicely. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3
#$&* course phy201 3/161:21 Question: `qprin phy and gen phy problem 4.02 net force 265 N on bike and rider accelerates at 2.30 m/s^2, mass of bike and rider
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Given Solution: `aA force Fnet acting on mass m results in acceleration a, where a = Fnet / m. We are given Fnet and a, so we can solve the equation to find m. Multiplying both sides by m we get a * m = Fnet / m * m so a * m = Fnet. Dividing both sides of this equation by a we have m = Fnet / a = 265 N / (2.30 m/s^2) = 115 (kg m/s^2) / (m/s^2) = 115 kg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qprin phy and gen phy problem 4.07 force to accelerate 7 g pellet to 125 m/s in .7 m barrel YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: m = 7 g ds = .7 m v0 = 0 m/s vf = 125 m/s dv = 125 m/s vAve = 125 m/s / 2 = 62.5 m/s dt = ds / vAve = .7m / 62.5 m/s = .0112 A = dv / dt 125 m/s / .0112 s = 11160.7 m/s/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The initial velocity of the bullet is zero and the final velocity is 125 m/s. If we assume uniform acceleration (not necessarily the case but not a bad first approximation) the average velocity is (0 + 125 m/s) / 2 = 62.5 m/s and the time required for the trip down the barrel is .7 m / (62.5 m/s) = .011 sec, approx.. Acceleration is therefore rate of velocity change = `dv / `dt = (125 m/s - 0 m/s) / (.011 sec) = 11000 m/s^2, approx.. The force on the bullet is therefore F = m a = .007 kg * 11000 m/s^2 = 77 N approx. ** STUDENT COMMENT: I did my answer a different way and came up with a number just off of this. I calculated 78 and this solution shows an answer of 77, but I am positive that I did my work right. INSTRUCTOR RESPONSE: The results of my numerical calculations are always to be regarded as 'fuzzy'. The calculations are done mentally and there is often no intent to be exact. This at the very least encourages students to do the arithmetic and think about significant figures for themselves. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qgen phy 4.08. A fish is being pulled upward. The breaking strength of the line holding the fish is 22 N. An acceleration of 2.5 m/s^2 breaks the line. What can we say about the mass of the fish? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T = m (a + g) 22 N = m (2.5 + 9.8) = 22 N = 12.3 * m M = 1.79 kg confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe fish is being pulled upward by the tension, downward by gravity. The net force on the fish is therefore equal to the tension in the line, minus the force exerted by gravity. In symbols, Fnet = T - M g, where M is the mass of the fish. (We use capital M for the mass of the fish to distinguish the symbol for mass from the symbol m for meter). To accelerate a fish of mass M upward at 2.5 m/s^2 the net force must be Fnet = M a = M * 2.5 m/s^2. Combined with the preceding we have the condition M * 2.5 m/s^2 = T - M g so that to provide this force we require T = M * 2.5 m/s^2 + M g = M * 2.5 m/s^2 + M * 9.8 m/s^2 = M * 12.3 m/s^2. We know that the line breaks, so the tension must exceed the 22 N breaking strength of the line. So T > 22 N. Thus M * 12.3 m/s^2 > 22 N. Solving this inequality for m we get M > 22 N / (12.3 m/s^2) = 22 kg m/s^2 / (12.3 m/s^2) = 1.8 kg. The fish has a mass exceeding 1.8 kg. STUDENT QUESTION I had trouble understanding this question to begin with. I am a little confused on why the net force equals an acceleration of 12.3. INSTRUCTOR RESPONSE F_net = M a = M * 2.5 m/s^2, as expressed in the equation F_net = T - m g so that M * 2.5 m/s^2 = T - M g. It is the tension, not the net force, that ends up with a factor of 12.3 m/s^2: T = F_net + M g = M * 2.5 m/s^2 + M * 9.8 m/s^2, which is where the 12.3 m/s^2 comes from. Nothing actually accelerates at 12.3 m/s^2, just as nothing in this system accelerates at 9.8 m/s^2. 9.8 m/s^2 is the acceleration of gravity so M * 9.8 m/s^2 is the force exerted by gravity on the fish. M * 2.5 m/s^2 is the net force on the fish. To not only pull the fish upward against gravity, but to also accelerate it at 2.5 m/s^2, requires a tension force of M * 2.5 m/s^2 in addition to the force required to overcome gravity. Thus the tension force is M * 2.5 m/s^2 + M * 9.8 m/s^2 = M * 12.3 m/s^2. STUDENT QUESTION So the T does not really factor out of the equation it is just known that it is greater thatn or less than the Fnet? INSTRUCTOR RESPONSE Fnet is M * 2.5 m/s^2. We know that T = M * 12.3 m/s^2. We know that since the string breaks T is at least 22 N. So M * 12.3 m/s^2 is at least 22 N, and M must be at least 22 N / 12.3 m/s^2 = 1.8 kg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `quniv phy 4.42 (11th edition 4.38) parachutist 55 kg with parachute, upward 620 N force. What are the weight and acceleration of parachutist? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): -Fnet = m * a -Fnet = 55 kg * -9.8 m/s/s = -539 N +Fnet = 620 N -539 + 620 = 80 N A = 80 N / 55 kg = 1.45 m/s/s Or T = m (a + g) 620 N = 55 (a + 9.8) A = 1.47 ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qDescribe the free body diagram you drew. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I drew a picture of a guy with a parachute with his mass beside him. I drew an arrow pointing up with 620 N and an arrow down with -Fnet = mass * a. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe weight of the parachutist is 55 kg * 9.8 m/s^2 = 540 N, approx.. So the parachutist experiences a downward force of 540 N and an upward force of 620 N. Choosing upward as the positive direction the forces are -540 N and + 620 N, so the net force is -540 + 620 N = 80 N. Your free body diagram should clearly show these two forces, one acting upward and the other downward. The acceleration of the parachutist is a = Fnet / m = +80 N / (55 kg) = 1.4 m/s^2, approx.. STUDENT COMMENT I am having a hard time still yet understanding conversions, Ex) kg*m/s^2 = N, these hard for me to compute. they are not as hard since zive been working with them but I am still having some trouble. INSTRUCTOR RESPONSE force = mass * acceleration, so the unit of force is the unit of mass * the unit of acceleration, i.e., kg * (m/s^2). We call this a Newton, but if you go back to the basic law you see why the basic unit is kg * m/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `quniv phy (4.34 10th edition) A fish hangs from a spring balance, which is in turn hung from the roof of an elevator. The balance reads 50 N when the elevator is accelerating at 2.45 m/s^2 in the upward direction. What is the net force on the fish when the balance reads 50 N? What is the true weight of the fish, under what circumstances will the balance read 30 N, and what will the balance read after the cable holding the fish breaks? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T = m (a + g) 50 N = m (2.25 + 9.8) 50 N = m * 12.25 m/s/s m = 4.08 kg When balance reads 30 N: T = m (a + g) 30 N = 4.08 (a + 9.8) A = -2.25 m/s/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Weight is force exerted by gravity. Net force is Fnet = m * a. The forces acting on the fish are the 50 N upward force exerted by the cable and the downward force m g exerted by gravity. So m a = 50 N - m g, which we solve for m to get m = 50 N / (a + g) = 50 N / (2.45 m/s^2 + 9.8 m/s^2) = 50 N / 12.25 m/s^2 = 4 kg. If the balance reads 30 N then F_net = m a = 30 N - m g = 30 N - 4 kg * 9.8 m/s^2 = -9.2 N so a = -9.2 N / (4 kg) = -2.3 m/s^2; i.e., the elevator is accelerating downward at 2.3 m/s^2. If the cable breaks then the fish and everything else in the elevator will accelerate downward at 9.8 m/s^2. Net force will be -m g; net force is also Fbalance - m g. So -m g = Fbalance - m g and we conclude that the balance exerts no force. So it reads 0. ** STUDENT COMMENT I totally messed this problem up. I still have a hard time knowing how to setup my problems, but I understand solution INSTRUCTOR RESPONSE There are usually numerous ways to set up a given problem. In the case of this problem you want to start with Newton's Second Law, which you did. Having calculated the net force you could have set it equal to 50 N - m g, which would have given you 12.5 N = 50 N - m g with solution m = (50 N - 12.5 N) / g = (50 N - 12.5 N) / (9.8 m/s^2) = 4 kg, very approximately The symbolic equation would be m a = T - m g with solution m = T / (a + g) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I am very confused by these problems ------------------------------------------------ Self-critique Rating:
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Given Solution: `a** Think in terms of net force. The net force on the fish must be Fnet = m a = m * 4.5 m/s^2. Net force is tension + weight = T - m g, assuming the upward direction is positive. So T - m g = m a and T = m a + m g. Factoring out m we have T = m ( a + g ) so that m = T / (a + g) = 22 N / (4.5 m/s^2 + 9.8 m/s^2) = 22 N / (14.3 m/s^2) = 1.8 kg, approx.. The same principles apply with the elevator. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Now that I have am equation I understand the problem a little bit more ------------------------------------------------ Self-critique rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!