course Phy 231 6/23 2:45 013. `query 13 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 #$&* OK ********************************************* Question: `qprin phy and gen phy problem 4.07 force to accelerate 7 g pellet to 125 m/s in .7 m barrel Your solution: Mass = .007kg Vf = 125m/s; v0 = 0m/s; ds = .7m 125m/s ^2 = 0m/s ^ + 2(a)(.7m) 15625m^2/s^2 = 1.4m a A = 11160.7m/s^2 Fnet = .007kg(11160.7m/s^2) = 78.1 N Confidence rating #$&*: 3
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `qgen phy 4.08. A fish is being pulled upward. The breaking strength of the line holding the fish is 22 N. An acceleration of 2.5 m/s^2 breaks the line. What can we say about the mass of the fish? Your solution: Net force = tension-gravity Mass*2.5m/s^2 = T-Mass(g) T = M 2.5m/s^2 + M 9.8m/s^2 T = M 12.3m/s^2 also tension cannot be above 22N 22N < M 12.3 m/s^2 M > 1.78 kg Confidence rating #$&*: 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. • 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 #$&* OK ********************************************* Question: `quniv phy 4.42 (11th edition 4.38) parachutist 55 kg with parachute, upward 620 N force. What are the weight and acceleration of parachutist? Mass = 55 kg Weight = 55kg(9.8m/s^2) = 539 N Air resistance = -620 N Fnet = 539N – 620N = -81N -81N = 55kg(a) A = -1.47m/s^2 ********************************************* Question: `qDescribe the free body diagram you drew. Your solution: The free body diagram has an arrow pointing up from the ground to the person; this represents the air resistance that is slowing the velocity of the parachutist. The other arrow is pointing down and is not as long as the first arrow because the force is not as great. Confidence rating #$&*:
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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.. Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `quniv phy (4.34 10th edition) A fish hangs from a spring balance, which is in turn hung from the roof of an elevator. The balance reads 50 N when the elevator is accelerating at 2.45 m/s^2 in the upward direction. What is the net force on the fish when the balance reads 50 N? What is the true weight of the fish, under what circumstances will the balance read 30 N, and what will the balance read after the cable holding the fish breaks? Your solution: M a = 50N – m g Ma+mg = 50N m(a+g) = 50N M = 50N / (a+g) M = 50N / (2.45+ 9.8) = 50N / 12.25m/s^2 = 4 kg 30N – 4 (9.8m/s^2) = -9.2 N A = -9.2N / 4kg = -2.3m/s^2 The balance will read 0 N because there is no mass on it. Confidence rating #$&*: 2
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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. ** Self-critique (if necessary): OK Self-critique rating #$&* OK ********************************************* Question: `qSTUDENT Self-critique rating #$&* ********************************************* Question: I had trouble with the problems involving tension in lines. For example the Fish prob. Prob#9 A person yanks a fish out of water at 4.5 m/s^2 acceleration. His line is rated at 22 Newtons Max, His line breaks, What is the mass of the fish. Here's what I did. Sum of F = Fup + F down -22 N = 4.5 m/s^2 * m(fish) - 9.8 m/s^2 * m(fish) -22N = -5.3 m/s^2 m(fish) m(fish) = 4.2 kg I know its wrong, I just don't know what to do.I had the same problem with the elevator tension on problem 17. Your solution: Net force = mass * acceleration or The force from tension plus the force from gravity Both of them have to be equal Confidence rating #$&*: 3
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Given Solution: `a** Think in terms of net force. The net force on the fish must be Fnet = m a = m * 4.5 m/s^2. Net force is tension + weight = T - m g, assuming the upward direction is positive. So T - m g = m a and T = m a + m g. Factoring out m we have T = m ( a + g ) so that m = T / (a + g) = 22 N / (4.5 m/s^2 + 9.8 m/s^2) = 22 N / (14.3 m/s^2) = 1.8 kg, approx.. The same principles apply with the elevator. ** Self-critique (if necessary): Ok Self-critique rating #$&* OK "