rasen58 Posted January 15, 2015 Share Posted January 15, 2015 As shown in the figure, a 2kg mass and a 3kg mass hanging from the massless pulley are released from rest. After the 3 kg mass has fallen 1.5 m, it is moving with a speed of 3.8 m/s. What is the rate at which the frictional force dissipates energy during this time interval? My solution: The 3 kg mass has Fg down and a Tension force up and the 2kg mass has a tension force to the right and a friction force to the left. With my coordinate system, the tension forces cancelled out, and I got Fnet = Fg - Ffr Ffr = Fg - Fnet Fg = 3 * 10 = 30 N Fnet = Mtotal * a To find the acceleration of the system, I knew Vi = 0m/s, Vf = 3.8 m/s, y = 1.5m Vf2 = Vi2+ 2ay Then by solving for a, I got 4.813 m/s^2 So now I know Fnet = (5 kg)*4.813 = 24.065 N And Ffr = 30 N - 24.065 N = about 6 N To find the dissipation of energy, power = work/time work = force*displacement The force here is the friction force = 6 N And I have to find displacement and time now. So to find time, looking at the 3 kg block y = Vit + 1/2 at2 Then solving for t gave me .7895 seconds. So in .7895 seconds, the 3kg block fell 1.5 meters, so now I need to find how much the 2kg block moved in that time period. So x = Vit + 1/2 at2 I use v initial = 0 m/s again (right?) and the same a that I found along with the t = .7895 And got 1.5 meters. So now to find the power dissipated, (6 N * 1.5 meters)/.7895 s = 11.39 Watts But the actual answer is 10 Watts. What did I do wrong? Thanks. Link to comment Share on other sites More sharing options...
studiot Posted January 15, 2015 Share Posted January 15, 2015 (edited) Yes I can confirm that the power is 10.19 watts, by another method. PE lost by 3kg mass = KE gained by system + Work done against friction. From the figures given 1.5*3*9.81 = 0.5*5*(3.8)2 + WF 8.045J Power = WF/t P = 8.045 *3.8/3 = 10.19watts. I agree with your acceleration and time calculations, again by a slightly different method. I am looking more closely at your method to see if I can spot the error. Edited January 15, 2015 by studiot Link to comment Share on other sites More sharing options...
imatfaal Posted January 15, 2015 Share Posted January 15, 2015 ... I am looking more closely at your method to see if I can spot the error. g = 9.81m/s^2 v g =10m/s^ Plus rounding force off too early 1 Link to comment Share on other sites More sharing options...
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