Question

PLEASE HELP! the graphs are attached

Procedure: Styles Editing 1. Click on the begin button and then click on the start button. You may change the Mass 1 and Mass 2 if it is necessary. Make sure the Mass 1 is always greater than the Mass 2. Record the mass values. Attach both graphs that you get using the simulation. Mass 1 = 115 kg Mass 2 105 kg 1 2. Draw free body diagram for each mass. m. m2 Focus

Answer 1

2) Free-Body diagrame T T M₂ 9 ap TA ma Mga mi mag mig for mass-1 For mass-2 mig o fa or, a= mi-m2 mitm2 3) Newton's Second law equation for mass 1 : mig-T = mia © Mass-2 (Ans) T-mag = mza G Here, a is the acceleration of the masses and T is the Tension of string: Adding equation (1) and @ we get, mig - m2 g = miat ma a g - This is the acceleration. (Ane) Now, putting the value of "as in equation (2) we get, Tension, T=mza+mag = mg g + or, , T= m2 a { mi-M2+ m, tma mitme or 2m, m2 T - This is the Tension of the mitm₂ string. Ans ) Now, putting the values of m. = 11569 m2 =105Kg - 9 months we get, acceleration, a = 0.445 m/s2 Tension of string, T= 1075.77 N es m, -mz mitm₂ ot Nazare

5) From First graph, we can see at t-o, position, 2=0 so; initial velocity is , U = 0 . Now, the equation of the parabolic graph is, x= hat? where a where a is the acceleration Now, we fee choose any arbitory point (1,2) to find acceleration from graph. we choose (0.75, 0.11 m) point. the position , x=0ill m. so, acceleration, a= 2 FROZOXIBA 2015 Parapha 2017)2 = So, Percentage error = So, at time, t=0.7 S a= 2X Ooll = 0.449 m/s2 (Ans) 0.449 -0.445 -X100% = 0.9 % Ans) 0.445 The soto slope of wat diagram (second graph) gives the acceleration, we take the slope of the line segment connecting the points (0,0) and (1.05,0.957) 50, acceleration , Q = f25,6-45m) 0.45m 0.45-0 AV At m/s? 1-0 0.0*15 → or, aa= 0.45 m/s2 'graph (Aus) 0.45-0.445x100-1. = 1•12% Ons) So, percentage error = 0.445