Group 1: Lab Activity # 7 Go to: Masses and Springs: Basics (colorado.edu) A. Solving for Hooke's constant In physics, Hooke's law is an approximation of the response of springlike bodies. unstretched 000 spring reell Hooke's Law: It takes twice Fspring=-kx as much force Ce 2X to stretch a Spring constant & spring twice as far. 2F Since Force is due to weight then -kx = mg Where x is displacement of the mass m is the mass k is a constant Choose stretch in the simulation. Use the default spring strength. Click all applicable boxes. Use the ruler for measurement. Mass ( in g ) Displacement x ( in cm) Graph the Force (weight) vs the displacement Calculate the slope. The slope should be equal to -k Slope = You can use this site for calculations: Slope and y-intercept of a Regression Line (Best Fit Line) Calculator (learningaboutelectronics.com)Since Force is due to weight then -kx = mg Where x is displacement of the mass m is the mass k is a constant Choose stretch in the simulation. Use the default spring strength. Click all applicable boxes. Use the ruler for measurement. (Gedo Cruz, Allen Dale) Mass ( in g ) Displacement x ( in cm) 50 g 8 cm 100 g 16 cm 250 g 41 cm Graph the Force (weight) vs the displacement Force [Weight) vs Displacement 2.5 2.0- 1.5 Force (in N) 1.0- 0.5- -. 0.0 0 10 20 30 40 50 Displacement (in cm) FORCE : 50 9 2.) 100 g f = m . a F = m . F : (50 g ) ( 9.81 m/s?) F - ( 100g ) ( 9.81 m/s " ) F = (0. 05 Kg ) ( 9. 81 mls ] ) F = (0. 1 kg) ( 9. B1 m/s = ) F = 0. 4905 N F = 0. 981 N F = 0. 491 N 3.) 290 g F = m q F = (250 q ) ( 9. 81 m /s ? ) F - (0. 25 kg ) ( 9.81 mls ? ) F = 2. 4525 N F - 2. 453 N