#$&* course PHY 201 101010 6:20pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aA force Fnet acting on mass m results in acceleration a, where a = Fnet / m. We are given Fnet and a, so we can solve the equation to find m. Multiplying both sides by m we get a * m = Fnet / m * m so a * m = Fnet. Dividing both sides of this equation by a we have m = Fnet / a = 265 N / (2.30 m/s^2) = 115 (kg m/s^2) / (m/s^2) = 115 kg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* 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: F=ma F=7g*125m/s^2 =875g*m/s^2 =.875N Confidence rating:: 2
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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): For starters on this problem, I see where I made my mistake in assuming the acceleration was given even though I know that acceleration is expressed as m/s^2 and not m/s. I also didn’t get the “picture” until I saw the solution of the barrel. I probably would have understood it better if it had been down a ramp but the problem was a barrel. I do now see how to solve this problem correctly. 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: F=ma 22N=m*2.5m/s^2 m=22N/(2.5m/s^2) m=(22kg/m/s/^2)/2.5m/s^2 m=8.8kg confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I didn’t use the formulas for tension force in this problem. I only used F=ma, which is a start to the problem, however, when dealing with this question, we have to also find the net force. Then we have, T = F_net + M g. After reading the solution to this question, I understand how to solve now. "