Question

didnt get the correct answer. please help

Required information NOTE: This is a multi-part question Once an answer is submitted, you will be unable to return to this part The double pulley shown has a mass of 15 kg and a centroidal radius of gyration of 160 mm Cylinder A and block Bare attached to cords that are wrapped on the pulleys as shown in the figure. The coefficient of kinetic friction between block Band the surface is 0.2. Consider that the system is at rest in the position shown when a constant force P-140 NIS applied to cylinder A 150 mm 15 5 kg Determine the total distance that block B moves before coming to rest. The total distance that block B moves before coming to rest is 3.95 m

Answer 1

Given dator, ma= 5 kg MA = 15kg = 180mm 420.2 P=1400 mc=15 kg distance moved by cylinder A is Sazim LLLLL222299999999 find the distanc moved by block A from the express ion. oes SA + = SA = 582 0.6m 250 150 draw the free body as shown in diagram of block figure below the Det .

The normal force acting on klock B is E Fn =0 рд - ид Cos Jo - 1519.81X до NB = 127.94 N friction for t= lenA= 0.2 x 127 eq f= 25.487 P find the work done by whole system. Olyz = PXSA & MAGSA - FSA - MAGSAY LEH Sino Ucun = 140 X 1 + 5895617 1 - 25.062 0.6 - 15x9:61 X 0.6 singo Oly2 = 129.61 J from work energy of final condition. principal of initial Ti + Oly2 T = kinetic = energy

ot 129.61=1 & MAUART I MBUAtla T Ž I c WC? 4 10 6 = ëmova mesos de 2462=ffa mount té mon amount te 2. I VB JAL .61 + x MAVAZ + Im + + I to \ Ic E 129.61 = √ MAUA? + IMAX ra? YAZ XUAR - ICAVAR YAZ was (mat me te reja 129,615 mat ma ro² YAZ 129.61 = mat mora? a mctc2 1 ra² 129.61- NI 5 + 15X.1502 15X.1627 t • 2502 252

129,613 Ź ( 5+ 5.847 61194) VA 2 129,61= 6. 272 VAZ / VA= 3.958 mis velocity af block A, UA= 8A UA VA VBS.:15. x 25 3,958 - 2.375 mie The angular wes speed of A - pulley co 3.45€ 15.892 rad 6s 125 When Cylinder A strike the the ground ground Then from inal principal of at Work and energy - Tot Oaya = To (žmog uge + 47.6.2%) + [- fso' - mog sa sin ga zo (ie finaly system come y system come to rest )

1 x 15 x 2.375 + 1x Cmcke 2 x wch - 25.987 So - 1589.818 Ull Solx sinso ¢2.304 + dx ( 15%.162).* 1 5.8322 - 99.602 SA' =0 (where SA= additional diston Ghovelles 90.429 = 99.60 2 SA' . by block B). SB'a ogolom find the total B block coming distance to the novelled by rest is welled by S.= SBt Sol So 0.6+ 0.9079 S = 1.5079 m