Question

if possible please write neat explaining how you got your answer so I could better understand problems like these in the future

A student hangs a 0.125 kg object on a spring, sets it into oscillation, and obtains the data for the position and velocity of the object as a function of time shown in the graphs below. 1001 0.80 +--A 0.60 Position (m) 0.40 + 0.20 +--- 0.004 0.0 0.5 1.0 1.5 Time (s) 2.0 2.5 1.01 0.5 Velocity (m/s) 0.5 +- S-1.0 - -1.0 - ---- -1.5 - 0.0 0.5 1.0 Times 2.5 (a) On the grid below, sketch the potential energy of the object-spring-Earth system as a function of time. Explicitly label any intercepts, asymptotes, maxima, or minima with numerical values, as appropriate. Potential Energy (J) 2.0 2.5 0.0 0.5 1.0 1.5 Time (s) Lahal anu inte

Answer 1

7 . Potential energy is given by . U = kx ? where it is distance of hanging mass from equillibrium position. From position us time graph we can see that equillibsium position = nint * max - *miy = 100 - 0.50 2 2 +0.50 = 0.25m +0.50 and An x= 0.75m amplitude A = A X max - min 1.00 -0.So 2 . 2 A = 0.25 m So Maximum potential energy. U= / KA² frequency of oscillation K= 4752m. = 4x (3.1972x 0.125 U = { x + x(3.1492 x 01125266-2952 When 0.15405625 Joule max

Potential enage will be minimum le. 2000 at times when it is at equillibrium position le at times f = 0.25, 0.75, 1.25, 175 sinner and is maximum at times . t=0, 0. So,lico, lSo, 20.. ДДДДД 0.25 0.5