course PHY 121 1/4 013. `query 13*********************************************
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Given Solution: A 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: ---- ********************************************* 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: ?? I’m not sure what to solve for or what the question is asking.. I think the question got cut off. I went back and couldn’t find it. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 – I know how to solve this. ------------------------------------------------ Self-critique Rating: ---- ********************************************* 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: The tension would have to be greater than 22N because the line broke. Fnet = m * a. The Fnet must be greater than 22N = m *a. The acceleration equals m*g, so 2.5m/s = ?? confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The 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. 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 an acceleration factor equal to 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got lost.. I am still working through how the acceleration was found since it isn’t very clear to me. I also don’t know what’s going on with these questions because most of the body of the question is missing. I checking my copy and paste and I copied what was there, so I’m not sure where the text was lost. Also, in the solution after this one, it states you can use 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 as well, where as in this specific problem the a = 2.5m/s^2, NOT 4.5. I was wondering about that discrepancy.