PHYS 125L Simulation Lab#S3

Name: Date: PHYS 125L Lab. # S2 Hooke’s Law and the Elastic Potential Energy ( Courtesy of the Department of Physics, University of Colorado, Boulder, Colorado ). Warning: Students are not allowed to share answers or results. Any simulation lab report that is deemed to be copied from another student or plagiarized will be assigned zero grade. All entries in the tables must be type written. All answers must also be typewritten. (No handwritten work will be accepted). Objectives: To verify Hooke’s law by determining the relationship between an applied force and the distance that a spring is stretched (measured from its equilibrium position). Hooke’s law states that the distance that the spring is stretched is linearly proportional to the applied force. The constant of proportionality is called the spring or force constant, k : F = k x Procedure: Open Hooke’s Law Lab using this link: https://phet.colorado.edu/sims/html/hookes-law/latest/hookes-law_en.html Part 1: Hooke’s Law with Two Springs in Parallel System Open the above link and click on the “Systems” play button. Once the simulation opens, choose the two parallel-springs system as shown in the following picture. Fill out the table on the next page and carry out the required calculations with the equations that are provided. Please answer the questions that follow your calculations thoroughly.

Here are the entries in the table: 1. Spring constant of the top spring, k t (N/m) 2. Spring constant of the bottom spring, k b (N/m) 3. Simulated Applied force, F a (N) 4. Simulated displacement that the springs are stretched, x s (m) 5. Calculated displacement that the springs are stretched, x c (m) k t k b F a x s x c % Difference k t x s k b x s F a ( k t + k b ) x s  40 0 400 10 .013 .013 0 5.2 5.2 .961 38 0 420 20 .025 .025 0 9.5 10.5 1 36 0 440 30 .031 .031 0 13.68 16.72 .987 34 0 460 40 .050 .050 0 17 23 1 32 0 480 50 .063 .063 0 20.16 30.24 .992 30 0 500 60 .075 .075 0 22.5 37.5 1 28 0 520 70 .088 .088 0 24.64 45.76 .994 26 0 540 80 .100 .100 0 26 54 1 24 0 560 90 .113 .113 0 27.12 63.28 .996 22 0 580 100 .125 .125 0 27.5 72.5 1 Calculated displacement that the springs are stretched, x c = F a ( k t + k b ) Questions: 1. Using Hooke’s law, F = k x , show that k = k t + k b when the two springs are in parallel. Note that the displacement x s is the same for both springs. Please show your steps clearly.

4 . Is F a equal to or close to k t x s + k b x s in each of the spring constants selected? Explain your answer. Hence for same displacement xs, Fa is same for both spring. Part 2: Hooke’s Law with Two Springs in Series System Using the same simulation, choose the two series-springs system as shown in the following picture. Fill out the table on the next page and carry out the required calculations with the equations that are provided. Please answer the questions that follow your calculations thoroughly. Here are the entries in the table: 1. Spring constant of the left spring, k L (N/m) 2. Spring constant of the right spring, k R (N/m) 3. Simulated Applied force, F a (N) 4. Simulated displacement that the springs are stretched, x s (m) 5. Calculated displacement that the springs are stretched, x c (m) k L k R F a k L × k R ( k L + k R ) x s % Difference x s x c % Difference 400 400 10 10 0 .05 .05 0 380 420 20 19.95 10 .100 .100 0 360 440 30 30.096 20 .152 .152 0 340 460 40 40.07 30 .205 .205 0 320 480 50 49.92 40 .260 .260 0 300 500 60 60 50 .320 .320 0 280 520 70 70.07 60 .385 .385 0 260 540 80 80.028 70 .456 .456 0 240 560 90 90.048 80 .536 .536 0 220 580 100 100 90 .627 .627 0

Calculated displacement that the springs are stretched, x c = F a ( k L + k R ) k L × k R Question: 1. Using Hooke’s law, F = k x , show that k = k L k R k L + k R when the two springs are in series. Note that the applied force is the same for both springs. Please show your steps clearly. A constant force →F is applied on spring 2. So that the springs are extended and the total extension of the combination is the sum of elongation of each spring. Alternatively, the direction of force could be reversed so that the springs are compressed. Part 3: Elastic Potential Energy of a One-Spring System Open Hooke’s Law Lab and click on the “Energy” tab as shown in the following picture. Fill out the table on the next page and carry out the required calculations with the equations that are provided. Please answer the questions that follow your calculations thoroughly. Here are the entries in the table: 1. Spring constant of the spring, k (N/m) 2. Simulated elastic potential energy, U s (J) 3. Simulated displacement that the spring is compressed (  ) or stretched (+), x (m)

Find this percent error: | 2 × coefficient − k | k x 100 = _________ %