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 HWQ200 Anon Sun 10/17 20:04 #lq303b-2 #q3218: For part a I am using the equation 0 = 0 + mg(deltaS) - f(deltaS) +k/2(delta S)^2. For the deltaS of friction I used 2.2 m and for the deltaS of the spring I used 0.23 m. What am I doing wrong - am I missing a component in the equation? 0 kkit Sun 10/17 20:28 What Δs did you use in the mg⋅Δs term? That term is usually written as mg⋅Δy. 0 adunca44 Sun 10/17 20:53 I forgot the deltaS there. I fixed it and got the right answer. 0 Add a response... HWQ199 jroark6 Sun 10/17 17:12 #lq303b-2 #q3218: For problem A, given that it starts and ends at rest, I set up the equation as 0 = 0 + mg(deltaS) - f(deltaS) +k/2(delta S)^2, with it rearranged to solve for f(deltaS), which I could undo to get mu_k. However, I cannot seem to get the right answer. I've tried not accounting for the work of the weight (mgDeltaS), flipped the signs on basically each term, moving over the tens place, etc - I keep getting wrong answers that make no sense for coefficients of friction. I don't know quite what I'm missing here; the plane is flat so there is no angle, so I don't need to get a component form of mg or any other force; starting and ending velocity is rest, so the 1/2mv^2 terms are zero. My best guess is there is another force besides gravity affecting the normal force, but I'm not sure what it would be, as there is no other force mentioned in the problem statement. 0 kkit Sun 10/17 17:18 Are you using different Δs values for the spring and friction? They are different. The spring is not attached to the block so it does not extend past equilibrium. 1 jroark6 Sun 10/17 17:21 I was using the same Δs value, so that is almost certainly the problem 0 Add a response... HWQ198 Anon Sun 10/17 14:49 For part B of number 2 on the post lecs, I don't know what I'm doing wrong. I've tried a couple different methods to no avail. My first try was to assume acceleration was constant, so I solved for Ax using the sum force x equation Fs(spring force) - f(friction) = MAx. Then I plugged all my values into the "no time" kinematics equation to try to solve for V2. My second try was setting net work equal to K2, (Fs-f)(x) = (1/2)mv^2 and solving for it that way but I ended up getting the same answer as the method I used in my first attempt. My third try was to set (1/2)mv^2 = -(1/2)kx^2, but that didn't work either. Any help is appreciated. Thanks! 0 Anon Sun 10/17 15:54 You are on the right track solving for K2 (since it has the variable we want to solve for, V2). K2=K1+total Work. kinetic energy at point B=kinetic energy at point A + total work You've already gotten several of the necessary terms from part 2a, so it's not that hard to solve from there. 0 kkit Sun 10/17 16:27 Your first method is not correct because the force of the spring changes with position. I wouldn't attempt to solve it that way. In your second method, you have an incorrect term for the work of the spring. It's not (Fs)(x) b/c the force of the spring changes with position. Your third method is closer to being a good one, but remember that friction is still operating when the block moves to the point where the block loses contact with the spring. 0 Anon Sun 10/17 17:07 Thank you guys- just got it done. 0 Add a response... WEQ197 Anon Sat 10/16 19:07 The "Energy of a Cart" video on the COE Equation learning page will not load. (It also says 0.00 length.) I tried to do the questions without it, but was making too many errors to proceed. 0 nhicks7 Sun 10/17 11:08 It looks like the source video might have been moved when we rearranged content. I’ll dig around to try to find where it went. 0 nhicks7 Sun 10/17 13:17 It should be fixed now! 0 Anon Sun 10/17 15:40 Thank you! 0 Add a response... LEQ196 Anon Sat 10/16 11:43 #lq304a-4 #q3233: Not sure what I am doing wrong on part A. I thought this was a simple question but I am evidently over simplifying it. Sense there is no friction force I simply set the work done by the spring, k/2(X^2) equal to .5(5.7/32.2)(v^2). What am I forgetting about here? 0 kkit Sat 10/16 12:08 There's an issue with your unit conversion (or lack of). I'm guessing you used g in ft/s/s but k in lb/in. 0 Add a response... LEQ195 Anon Fri 10/15 19:24 #lq303b-1 #q3216: Is the Mg supposed to mean milligrams for the mass? 0 kkit Fri 10/15 21:40 No. Look up the SI prefixes. 0 Add a response... HWQ194 cbock Fri 10/15 13:29 #lq304b-1 #q3234: I think the times are messed up for this as it says it is due on Monday instead of opening on Monday... 0 nhicks7 Fri 10/15 13:48 Ah! I was mixing up dates for pre and post lecture questions when setting it up. Thanks for the heads up! 0 Add a response... HWQ193 Anon Thu 10/14 21:20 #lq302b-2 #q3211: On part C of the second problem, I don't know what I'm doing wrong. I tried using the kinetic energy equation for calculating work. Since the object started at rest, the starting kinetic energy should be zero, so I should be able to just solve for v with the equation, W = 0.5mv^2. I tried using a different method as well. I found the total sum of forces on block A--gravity on block A parallel to the incline - (tension from block b + friction)--using the equation: sin(theta)ma9.81 - (mb9.81 + cos(theta)ma9.81uk), then I divided the resulting value by the mass of block A to find the acceleration. I plugged the acceleration of block A alongside the distance into the equation for projectile motion resulting in this equation: 2.25 = a*0.5*t^2, and I solved for t to find the time, which I multiplied by the acceleration to find the exact same value that my previous method gave me. I've been troubleshooting this problem for upwards of an hour and can't seem to find out where I've gone wrong, especially since the two methods I used don't share a common denominator other than block A's mass. 0 kkit Fri 10/15 8:27 I think you were making this harder than it needs to be. You don't need to solve for acceleration or time. In B you solved for the total work of the system. Since the system starts from rest the final kinetic energy of the system is equal to the total work done (B answer). You can calculate the final speed from the final total kinetic energy. 0 Anon Fri 10/15 10:48 I did do that. I only solved for acceleration and time to check the answer that I got from solving for velocity from the kinetic energy equation. I described both of the methods, the kinetic energy method and the acceleration method, in my question because they both yielded the same answer despite not having any common factors where I could've made a mistake. The total work done for my part B is 130. J, the mass of block A is 17.0 kg, the mass of block B is 7.68 kg, the angle of the incline is 64.7 degrees, and the kinetic friction coefficient is 0.250. With the kinetic energy method, I'm inputting the equation ((130*2)/17)^0.5 into my calculator to receive 3.91 m/s as the answer With the acceleration method, which I did in order to check where I went wrong in the kinetic energy equation, I'm inputting the equation: ((sin(64.7)*17*9.81)-((7.68*9.81)+(cos(64.7)*17*9.81*0.25)))/17 to find the acceleration, which is 3.39 m/s/s, and then inputting the equation: (((2.25*2)/3.39)^0.5)*3.39 to find the time and multiply it by acceleration. It also resulted in a value of 3.91 m/s I don't know what I could be doing wrong in either method--especially the kinetic energy equation--to have both methods yield the same, incorrect answer. Could you let me know what mistakes I'm making in each method? 0 nhicks7 Fri 10/15 11:06 Remember that because the two objects are connected, you're not just looking at the mass of block B. The reason you're getting the same incorrect answer using different methods is because you're not properly accounting for the mass of block B with either method. 0 Anon Fri 10/15 11:13 Thank you so much! 1 Add a response... HWQ192 lhochste Wed 10/13 15:23 #lq302b-2 #q3211: How do you find the force of tension in part A? So far I have been trying to find it through the F=ma equations and then using that value in the work equations to find the work for A or B, but this has not been successful and I am stuck. 0 kkit Wed 10/13 16:44 The activity in class today showed that the value of the tension is not needed. You can review the notes on the website. 0 Add a response... LEQ191 Anon Wed 10/13 14:59 #lq302b-2 #q3211: I am not sure what I am doing wrong on part B. To calculate my Work on block B I set up the equation. Wb=2.25(T-W) where T=tension and W=weight. To calculate my work on block A I set up the equation Wa=2.25(T+F-W) Where T=Tension F=friction force and W=weight. I then subtracted work b from work a to get the total work done. I got Wab= 2.25(f+weightb-weighta). What is wrong with my strategy here? 0 kkit Wed 10/13 16:50 The Wb equation is correct. At least two of the signs in the Wa equation are incorrect. Also, you should add Wa and Wb to get the total work. 0 Anon Wed 10/13 17:00 Do I need to account for the total displacement by block A, or just the vertical displacement of block A? 0 Anon Wed 10/13 19:33 I reset my equations to get Wa=2.25(-T-f+wsin(58.5)) and Wb=2.25(T-w). I added these together to get Wab= 2.25(-f+weightasin(58.5)-weightb) I am still getting the wrong answer. 0 kkit Wed 10/13 20:18 What you wrote looks correct. What's your value of f? 0 Anon Wed 10/13 20:24 That was my problem. I am not sure what I did so wrong but my friction force was off by a factor of two. I got it now, thank you! 0 Add a response... GQ190 cbutner1 Tue 10/12 22:39 #lq301b-1 #q3222: For part D it wanted my answer to be negative because the work done by weight was larger than that done by Elsie. I'm just confused because the problem said the cart moved up the slope. Is this just from my values being weird? I thought if work was negative the cart would be going down the slope. 0 nhicks7 Tue 10/12 22:53 If I throw a ball in the air, while it is traveling upward toward its apex, gravity will be the only force acting on the ball. In that case, because gravity is acting in the direction opposite motion, the work gravity does in negative. However, because it is the only force acting on the ball over that motion, that also means the total work done while the ball goes toward its apex is also negative--but it still goes up to its apex, right? But what does happen is the ball slows down while negative work is done on it (as you'll cover in lecture tomorrow and Friday). All this means is that the speed slows down. But you were not told anything about the initial or final motion of the crate, so there's nothing about the problem that is inconsistent. 0 Add a response... GQ189 abonsted Tue 10/12 21:37 It says I received at 65.8 for my Pre Lecture 3-1 even though I believe I completed everything. Can you confirm and/or let me know if I didn't see a part of the Pre Lecture? 0 nhicks7 Tue 10/12 21:53 Things got messed up when I removed part B from problem 3. I fixed it for you. 1 Add a response... HWQ188 Anon Tue 10/12 21:37 #lq301b-1 #q3222: I have no idea what I'm doing wrong with 1c. I've tried it so many different ways and gotten the wrong answer every time and I just have no idea where to even start now. 0 nhicks7 Tue 10/12 21:50 What is the direction of the normal force? What is the angle between the normal force and the direction of motion? 0 Add a response... LEQ187 Anon Tue 10/12 20:04 #lq301b-1 #q3222: I'm stuck on part b of the post-lecture question. I know that the work applied by Elsie on the cart is just the force times the distance but I'm unsure on how to account for the angles. 0 bmilstea Tue 10/12 20:23 I'm stuck on part D right now. I thought that you would find total work by just adding all of the work done on the cart together. Which in this case would be the work of Elise and the weight of the cart. This does not seem to be correct but I have no clue on how else to solve it. 0 Anon Tue 10/12 20:24 How did you do part B? 0 nhicks7 Tue 10/12 20:32 For part B: even though it gives you an x-y coordinate system to use, you are free to orient your perspective however you would like to calculate each part of the problem. The cart's motion is up the incline, so think about finding the component of the applied force that's in that same direction. The angle of the force is relative to the top of the cart (which is parallel to the incline, and therefore in the same direction as the motion). It's a bit tricky to think through! For part D: You're right, if you add up the work done by each of the components (Elise and weight), you'll get the total work done on the cart. Be careful to use your unrounded values! Because the two work values are pretty close, the answer is a smaller order of magnitude. So when you get that final answer and need to go to 2 sig figs, if you use rounded intermediate values, you lose crucial information. This is why we always recommend keeping the unrounded values (out a few sig figs past your answer), even if one of those values is an answer to a subpart of a question. 0 nhicks7 Tue 10/12 20:34 If my explanation about part B is still somewhat unclear, remember, the only thing that matters with work is the portion of the force that's in the same direction as the displacement. So, you can basically ignore the angle of the incline. 0 nmoore33 Tue 10/12 20:37 that helped thank you 1 Add a response... HWQ186 Anon Tue 10/12 17:15 #lq302a-3 #q3199: I assume these bodies are not connected? I don't have an image of the system. 0 nhicks7 Tue 10/12 17:34 Yeah, there isn't an image for the problem. It is fair to assume that they are not connected (we'll think about this context again in EF 158 when we learn about gases). If they were connected, whatever connected them would be experiencing changes in energy that would significantly complicate the problem. 1 Add a response... HWQ185 Anon Tue 10/12 16:12 #lq302a-3 #q3199: I'm not sure what I am doing wrong on part B. I am using W=Kf-Ki and using the W value given and the value from part A for Ki to solve for Kf. Is there something wrong in my method? 0 nhicks7 Tue 10/12 17:35 No, your first answer was totally correct, just not the right number of sig figs! 0 Add a response... HWQ184 Anon Mon 10/11 15:55 #whw204-1 #q3200: Do we need to show 2 or 3 sig figs in our answers? The angle only gives 2 but all of the answers use 3. 0 nhicks7 Mon 10/11 15:57 The answers provided in the homework are only intended to help guide you as to whether you're using the correct process. Use the appropriate number of sig figs based on the sig fig rules on the work you submit. 0 Add a response... HWQ180 Anon Sat 10/9 16:48 #whw204-1 #q3200: On part C , I am really confused on how to get the maximum velocity because I did Vterm= sqrt(2mg/PACd) and I got 367 as an answer. Can you give me a hint to what I could be thinking about or doing wrong? 0 Anon Sat 10/9 17:46 I'm having much the same problem - using either: v_ter = mg/b or v_ter = sqrt(smq/pACd) gives the wrong answer 0 nhicks7 Sat 10/9 18:08 Recall what I said in class when I introduced terminal velocity and during the review yesterday—do not rely on those formulas. Those only work in the context of free fall when the only two forces acting on the object are gravity and drag. The key concept is that the drag force varies with velocity and therefore as your speed increases, your drag increases and your overall acceleration decreases. Terminal velocity occurs once that acceleration reaches zero. So how have we been solving for acceleration this entire module? Doing the 8-step process! Draw your FBD and KD, write your 2nd law equations (ΣF=ma), but in this case we set acceleration to 0. Because one of your terms is drag and includes velocity, you then solve for the velocity that makes that equation true. So, to reiterate, don’t rely on the vT=mg/b or the quadratic equivalent—the math will change as the context changes. 0 cweems3 Mon 10/11 15:32 For part C, I set up my sum of forces equation as Wsin(theta)-Wcos(theta)(rho_k) - bv = 0 I then got bv = 52.234 and solved for v by dividing by the drag coefficient that I got in part B. Anything I am missing? Does b not equal that C_D that I got in part B? 0 cweems3 Mon 10/11 15:38 Nevermind. I figured it out using the formula for pressure drag instead of bv. 1 Add a response...