Question

Problem 1: The system shown is at rest. The cable is massless but the pulley is not. Use Newtonian mechanics and Second Law evaluations to solve the problem. a.) When the masses are released, what is the acceleration of block B? b.) What is the velocity of block B when it has moved 2 meters. Jo = 0.7 kg*m2 R = 0.2 m Ma = 7 kg MB = 20 kg B u = 0.2 (all surfaces) 45° 3001

Answer 1

solution: MA = 7kg, MB=20kg, To = 0-7 kg - w°, К = 0.2 т м = 0 .2. ед – 45°, ер = 20° FNA = Mag Costa FA = FЈА И — MMA-9 Сед ИА 9 Сөр ТА – НА 9 ѕіме,- ИМА 5 Съ0 - Маа Magsing AFFA овч Ba in FNB FNB = MB 9 GOSOR FfB Meg cost Fe0 = AFN) = ИМВ 9 СӨВ OMBg sino " MBg sino - T = MM Bg so = MB a and

ol 1 (To-TA) R = Jo x since string is not slipping TB 2= ♡ - TB-TA = Toa using the above equations, MBgsing-Magsinga-mg cos OA MA-ugos 0 MB za lent MA +MB) a. 9{MB SNOB-MA sinda-.M (MACODA + MBOSOR) To at MA + MB . 9-81/20 sin30 – 7 sin 45 -0.21760345 + 20 60330')} a= 0.7 0:22.7 7+20

a= 0.13143 m/s2 6) using the equation Uf2_v; ²= zas vito, vp = [zas]" Vf=[2x0.1314342] = 0.725 m/