Question

6. The system consists of 60-lb and 20-lb block A and B, respective, and 5-lb pulleys C and D that can be treated as thin disks. Assume that the cord does not slip on the pulleys, and neglect the mass of the cord: a) Show clearly all individual terms for the potential energy and kinetic energy b) Determine the speed of block A after block B has risen 5-ft, starting from rest. c) Determine the tension in the cord at block B while the block is in motion. 0.5 A D 0.5 - B 888

Answer 1

ILL 5012 @ Kinematics. The speed of block A & B. can be relate using the position coordinate equation. SA +25B=0 DSA +2 ASB=0 DSA +215)=0 -- ASA=-10 ft=10 ft () Taking time destrative of eq no , SA VAT 2VB=0 SB a los VB=-O-SVA) 6 Rotential Energy Datrem is set at fixed pieley Cooghen blocks of A B (pieliey D) are at their initial position, their centres of gravity are located at SA & SB, their initial gravitational potential energies que ECOSA) (-205B) & -sse). When block B (pulley ) rises 5ft , block A descends 10ft, thus the final position of blocks A & B(pelley D) are, (SA+10) ft &($8-5) Ft belowo datem. Hence, their respective final gravitational p.es que 660 (SA+103), 6-20 (56-5)] & Esc56-55). Thus, the mitial & final p.e's are, Vi=-60 SA - 2058 - 598 = C-GOSA -2558) va -GoCSA +10) - 20( 58-5) = 5(58-5) V2.=(-605A -255B-475) Kinetic Energy The mass manent of mentia of pulley about its mass (Xoos) . IG= 0.9941 slag.ft? Since pulley is solis without slipping, wipe VB VB 0.5 = 2 V3 =2(-0-514)=-VA Pulley C rolades about fixed point, Hence, wc-va = 2VA Since system initially seet The final KE is given by [To=0 T2 = IMAVA + 1 move + 1/2 mb vat LIGUB + ZIG WE² ） centre is IG- the initial Kinetic energy rs

32.2 7 60 32-2 T2 = Print $ (201) (-0.500*+ Ź 0.5VA)? + { (0.01941) Gvad (0.9841) (2VA) 2 T2 = 1.0773 VA conservation of Energy using eqn, Titvi = 12tV2 0+ (-60 SA -2558) = 1-0773V+ (-605A -2858-475) VA=21-0 ft/see d