Error Propagation Of Kinetic Energy
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Community Forums > Science Education > Homework and Coursework Questions > Introductory Physics Homework > Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors Dismiss Notice Dismiss Notice Join Physics Forums Today! The friendliest, high quality science and math community on the kinetic energy uncertainty formula planet! Everyone who loves science is here! Uncertainty of Kinetic Energy Jul 7, 2013 #1 bysons how to find measurement uncertainty for kinetic energy 1. The problem statement, all variables and given/known data An object of mass m = 2.3 ± 0.1 kg is moving at a speed of v percentage uncertainty formula for kinetic energy = 1.25 ± 0.03 m/s. Calculate the kinetic energy (K =(1/2)mv2) of the object. What is the uncertainty in K? 2. Relevant equations Δz = |z| ( Δx/x + Δy/y ) - Multiplication Δz = n (xn-1) Δx - Power 3. error propagation physics The attempt at a solution k=1/2 mv2 (power) Δv = 2(1.25)1(0.03) =0.075 (multiplication) Δk = k(1/2) (0.1/2.30 + 0.075/1.25) =0.09 1.80 ± 0.09 kg*m2/s2 Just wondering if this is correct or if I have gone about this wrong. I'm I correct in multiplying K by 1/2 in the last step? bysons, Jul 7, 2013 Phys.org - latest science and technology news stories on Phys.org •Game over? Computer beats human champ in ancient Chinese game •Simplifying solar cells with a new mix of materials
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•Imaged 'jets' reveal cerium's post-shock inner strength Jul 7, 2013 #2 haruspex Science Advisor Homework Helper Insights Author Gold Member bysons said: ↑ Δv = 2(1.25)1(0.03) =0.075 More accurately, that's Δ(v2). (multiplication) Δk = k(1/2) (0.1/2.30 + 0.075/1.25) How do you justify the inclusion of the factor 1/2? I don't see that in the equations you quoted. Let's say we replace the 1/2 with an unknown, a: k = a m v2 Δk/k = Δa/a + Δm/m + Δ(v2)/(v2) Knowing that a = 1/2, no error, what would you write for Δa? haruspex, Jul 7, 2013 Jul 8, 2013 #3 bysons Okay I did the work right though I just wrote delta V instead of delta v^2? 1/2 is just a constant in the kinetic energy formula. I just treat it as something still being multiplied? bysons, Jul 8, 2013 Jul 8, 2013 #4 haruspex Science Advisor Homework Helper Insights Author Gold Member bysons said: ↑ Okay I did the work right though I just wrote delta V instead of delta v^2? Yes. 1/2 is just a constant in the kinetic energy formula. I just treat it as something still being multiplied? No. Think about what I wrote before. 1/2 is a precisely known constant. If z = xy and x is precisely known, what do you get for Δz/z? haruspex, Jul 8, 2013 Jul 10, 2013 #5 bysons Δz/z = Δy/y ? bysons, Jul 10, 2013 Jul 10, 2013 #6 haruspex Science Advisor Homework He
Error Propagation For a function of several variables, $f
Propagation Of Error
= f(a,b,c,…)$, the propagation error in a measurement of error propagation rules $f$ is$\sigma^2_f = (\frac{\partial f}{\partial a})^2 \sigma^2_a + (\frac{\partial f}{\partial b})^2 \sigma^2_b + (\frac{\partial relative uncertainty f}{\partial c})^2 \sigma^2_c +…$In your next lab, you will be introduced to the concept of kinetic energy, $K = (1/2)m v^2$ where https://www.physicsforums.com/threads/uncertainty-of-kinetic-energy.700418/ $m$ is the mass of the object and $v$ is its speed having the following values: $m = 4.7 \pm 0.5 (kg)$ $v = 0.4 \pm 0.2 (m/sec)$ Using the method of propagation errors, what is the kinetic energy and its uncertainty? Make certain you http://www.show-my-homework.com/2015/06/error-propagation-in-measuring-the-kinetic-energy.html carefully show all your work. The total kinetic energy is$K = (m*v^2)/2 = (4.7*0.4^2)/2 =0.376 Joules$Differentiating twice the equation of Kinetic energy we obtain$d K = (d K/d m)*d m + (d K/d v)*d v$$d K^2 = (d K/d m)^2*(d m^2) + (d K/d v)^2*(d v^2)$ We know that$(d K/d m) = (1/2)v^2 =0.5*0.4 =0.2$and$(d K/d v) = m v = 4.7*0.4 =1.88$with the errors in mass and speed given as$d m =0.5 kg$ and $d v =0.2 m/s$ Therefore$(d k)^2 = 0.2^2 * d m^2 +1.88^2*d v^2 =0.04*0.25 +3.5344*0.04 =0.151376$ Related valentin68 Search for: Recent Posts 1D two particles system (Lagrangian) Car Tire Presssure. Zenner Regulator Coaxial Cylindrical Capacitor Swinging Pendulum (Quantum) CDs and DVDs Acoustic Sound Intensity Potentiometer and Errors Copyright 2012-2016 @ Show My Homework Theme by Colorlib Powered by WordPress
Error Propagation Projects Lab, University of Utah print Very often we use our physical measurements as a means to some computational end. We may, for example, use measurements of mass and velocity to calculate kinetic energy, or temperature and http://www.che.utah.edu/~tony/course/material/DataAnalysis/10_error_propagation.php pressure to calculate molar volume. While we should have a fair grasp on the uncertainty inherent in our physical measurements, we are also interested in bounding the uncertainty in those calculated values. Methods of error propagation allow us to translate the error in independent variables into the error within the dependent variables of our functions. Introduction Error Propagation for Arbitrary Functions Analytical Method for Error Propagation Numerical Method for Error Propagation Monte Carlo error propagation Method for Error Propagation Error Propagation Example Introduction: Let's take a very simple example (This example will illustrate the difference of two variables, but the results for error propagation in addition are the same). Say we wanted to know the weight of a liquid in a container. We take the weight of the empty container, w1, then fill it with our liquid and take the weight again, w2. Each weight, w1 and w2, for kinetic energy comes with a range of uncertainty, ±e1 and ±e2 respectively (always at some confidence level). In this example, these errors depend on the precision of the scale we used. Of course, the weight of the liquid, w0, should be the difference, w2 - w1, but what is our uncertainty in w0? At first glance, we might expect the range of w0 to be between the maximum and minimum values possible if we take our measured weights at their extremes: (1) (2) making the uncertainty (error) in w0: (3) . However, in reality, this value of e0 is too pessimistic. Think of tossing two six-sided dice and adding the resulting numbers together. With individual die, we have a 1 in 6 chance (16.7%) to get each number, including the extremes of 1 and 6. However, when we add the die together, the numbers at the extremes of that calculated value become less likely. For the extremes in the calculated value of 12 or 2, the probability drops to (1/6)*(1/6)=2.78%, while the mean value of 7 remains at 16.7%. To illustrate this concept with our liquid weight example, let's assume w1 = 1 g, w2 = 2 g, and the error associated with both measurements is ± 0.1 g. For simplicity and illustration, assume the
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