Question

PHY 202 CHAPTER #11 - PRACTICE PROBLEM #1 VERTICAL SPRING Y_O new y_0 F = -kx Hang a 0.200 kg block off of a spring. a.) The spring stretches 1.09 cm from its equilibrium point. What is the spring constant? ANS: 1.80 x 102 N/m E = KA? b.) Pull the block a further 1.60 cm down. The block will oscillate with what amplitude, and what total energy? ANS: 1.60 cm, 0.0230 J
Hang a 0.200 kg block off of a spring.

Answer 1

block mass fm )=0.200 kg a) spring stretched distance (x ) = 6.09 cm 2= 0.0109 m Spring constant (k): kxam . mg mg 0-200 * 9.81 k X 0.0109 2 k = 180 = 1.80 X10 wilm b) further distance (AP) - 1.60 cm A = 0.0160 m. Total energy E = 1 KA² F = x180x0.0160 = 0.023. J 2

c) period (7) 0 200 T = 21 21 = 211 le 180 T = 0.209 T=0.210s (1-0-240 frequency (f); f=t 11 4.77 Hz T 0.210