GROUPW Simulation 1 Group - 10 questions Start by going to this webpage for the Hooke's Law simulation at: Hooke's law Simulation link Select "Intro". Check all five boxes on the right hand side (applied force, spring force, displatementr equilibrium, values). Play around with the red slider control for the applied force. 0 Applied Force C] Spring Force ' C] Displacement % 2 C) Equilibrium Position 5. O Applied Force Spring Force O Displacement O Equilibrium Position Values Spring Constant 1: 200 N/m Applied Force 1: ON 100 500 1000 -100 100 4 Hooke's Law PHET = Intro SystemaApplied Force Spring Force Displacement ON:ON Equilibrium Position Values 0.01 Spring Constant 1: 200 N/m Applied Force 1: ON 100 500 1000 -100 100 Hooke's Law PHET = Intro Systems energyQUEWION1J 01 Choose one '10 points How does the magnitude and direction of Spring force compare to the applied force? 0 They have the same magnitude and same direction "._ They have the same magnitude but opposite directions 0 The spring force has less magnitude but same direction as the applied force. 0 The spring force has a greater magnitude and opposite direction as the applied force. QUESTION 1.2 V Q2 Choose one . 10 points Set the applied force to +100 Newtons. Play with the blue slider which adjusts the spring constant. See the effect of a higher and lower constant on the equilibrium position and forces. What is the effect of the spring constant on the spring. O A higher value for k result in a stiffer/less stretchy spring O A higher value for k results a less stiff/more stretchy spring O A higher value of k has no effect on the stiffness/stretchiness of the spring.QUESTION 1.3 V Q3 Choose one . 10 points Return the value of the spring constant back to 200 N/m. Change the value of the applied force to any number of your choice. On a scrap piece of paper, make note of this applied force and the displacement of the equilibrium position(green arrow). Do this several times so you can start to see a pattern in their relationship. What is the relationship between the applied force and the green displacement vector for a constant k? They are directly proportional (Linear relationship) They are exponentially proportional (quadratic relationship) They are inversely proportional ) You are a mean teacher and are trying to trick me. There is no relationship.QUESTION 1.4 Q4 Choose one . 10 points In 1678 English Physicist Robert Hooke published that "As the extension, so the force". He established that most solids behave (at times) with elastic properties; even very "inelastic" materials like steel will behave elastically under large loads. . In short he resolved that Fs = -kx where: . Fs is a "spring force" or "restoring force" (as the spring tries to return to its original or unloaded form) (Units: N) . k is the "constant of elasticity" or basically number that describes how elastic or stretchy a material is. (units: N/m) . * is the elongation or the deformation of the spring. Basically the difference in length of the spring when stretched from its unstretched length. (Units: m) . The negative sign indicates that the Spring Force is in the direction opposite that of the displacement (elongation). Look at the values of displacement and applied force you found for question 3. Do your values match Hooke's equation? Yes No (you did something wrong. Go back and figure out what went wrong and try this question again!)QUESTION 1.5 V 05 Choose one . 0 points Now let's hypothesize on the effect of having two springs working in parallel. (See picture below) Assume each spring has a constant of elasticity of 400 N/m when a force is applied. If you used Hooke's law to find the combined spring constant (k) of the combined springs, how do you think it would compare to the spring constant of of just one of the springs? Think about it and make your best guess. The combined spring constant will halved to 200 O The combined spring constant will stay the same at 400 O The combined spring constant will double to 800QUESTION 1.6 06 Choose one . 0 points Now let's hypothesize on the effect of having two springs working in series. (See picture below) Assume each spring has a constant of elasticity of 400 N/m when a force is applied. If you used Hooke's law to find the combined spring constant (k) of the combined springs, how do you think it would compare to the spring constant of of just one of the springs? Think about it and make your best guess. The combined spring constant will be halved to 200 O The combined spring constant will be the same at 400 O The combined spring constant will be doubled to 800QUESTION 1.7 07 Choose one . 10 points Click on the "systems" icon at the bottom center of the PhET simulation. You will now test your predictions. Applied Force Spring Force Displacement > 100 N 100 N Equilibrium Position Values 0.500 m Spring Constant 1: 200 N/m Applied Force 1; 100 N 100 500 1000 -100 100 Hooke's Law PHET= Intro Systems EnergyCheck all the boxes on the right. Make sure your 2 springs are in parallel like in Q5. Set them both to a spring constant of 200 n/m. Use Hooke's Law (F=-kx) to solve for the spring constant of both springs combined. Do this by applying force and finding the resulting displacement. Use this force and displacement to find the spring constant(k) of the combines springs. What is the new spring constant for the 2 springs in parallel? ON ON Applied Force Spring Force O Tote O Components al Displacement > Equilibrium Position Values 0.000 m Top Spring: 200 N/m 200 400 600 Applied Force: ON -100 100 Bottom Spring: 200 N/m D 200 400 600 Hooke's Law 110 Enemy PhET= Inin Swimama\fQUESTION 1.8 08 Choose one . 10 points Change your spring arrangement to series by clicking the button with 2 springs connected end to end on the right side of the simulation. Make sure your 2 springs are in series like in Q6. Set them both to a spring constant of 200 n/m. Use Hooke's Law (F=-kx) to solve for the spring constant of both springs combined. Do this by applying force and finding the resulting displacement. Use this force and displacement to find the spring constant(k) of the combines springs. What is the new spring constant for the 2 springs in seriesl? Applied Force Spring Force O Total ON ON O Components Displacement ) Equilibrium Position Values 0.000 m Left Spring: 200 N/m Right Spring: 200 Nim Applied Force: ON 200 409 200 100 100 0 100 Hooke's Law Intro Systems Energy PHET =@ Applied Force ) Spring Force O Total ON ON O Components Displacement Equilibrium Position Values 0.090 m Left Spring: 200 N/m Right Spring: 200 Nim Applied Force: ON 200 200 100 -100 100 Hooke's Law Intro Systems Energyy N PHET = 100 N/m 200 N/m 400 N/m 4,000 N/mQUESTION 1.9 09 Choose one . 10 points Click "Energy" button at the bottom center of the simulation. Change the spring constant on the one spring to 400 N/m. Put a checkmark in the box next to "values" on the right side of the sim. Use the green slider to adjust the displacement of the spring and watch what happens to the value of the potential energy. On a scrap piece of paper, write down the vaults of potential energy for a displacement of -1, -0.5, 0, 0.5 and 1. How is the sign (positiveegative) of the potential energy compare to the sign of the displacement? Applied Force Spring Force O Total ON ON O Components Displacement - Equilibrium Position Values 0.000 m Left Spring: 200 N/m Right Springs 200 Nim Applied Force: ON 200 409 600 200 100 100 100 Hooke's Law Intro Energy PHET=O Bar Graph O Energy Plot O Force Plot Potential Energy Energy - O Applied Forge O Displacement O Equilibrium Position O Values Spring Constant: 100 N/m Displacement: 0.000 m 100 200 300 400 D Hooke's Law Intra PhET = Spaatoms Energy They always have the same sign They always have opposite signs Potential energy is always positive Potential energy is always negativeQUESTION 1.10 w Q10 Choose one . 10 points Using the data you wrote for Q9, what is the relationship between the magnitude (absolute value) of the displacement and the potential energy. O No relationship Directly proportional (linear) O Exponentially proportional (quadratic) O Inversely proportionalGROUP 2 Simulation 2 Group . 2 questions Now lets go to a different simulation: Go to: Mass and springs simulation link Press play. Select "Stretch" On the top right add checkmarks to "Unstretched Length" and "Resting Position". Do not adjust the spring strengths.QUESTION 2.1 Q11 Choose one . 10 points Using Hooke's Law (F=-kx), the labeled masses (50g =>250g) and ruler in the simulation, find the spring constant k of spring #1. (Hint: The force applied by adding the masses is the weight of that mass. Weight(newtons)=mass(kg)x9.8m/s2) Be careful of units. What is the spring constant (k) of spring #1? O 0.6 O 4.2 O 6.0 O 7.3 9.8QUESTION 2.2 Q1 2 Choose one - 10 point: Now that you know the spring constant of spring #1 , use it to find the mass of the blue block. What is the mass of the blue block? 0 0.050 kg 0 0.100 kg 0 0.1?5 kg 0 0.245 kg