1. The problem statement, all variables and given/known data http://www.sumoware.com/images/temp/xziltxnrjlckqjqs.png [Broken] Two blocks ( A and B ) are linked by rope through three identical pulleys C, D, and E. Pulley C and E are in stationary, while D is pulled down so its constant velocity 1 m/s If vA = 4m/s when A goes down 2 m, determine the velocity and acceleration also the direction of block B when A goes 2 m down Choices : a) 4 m/s and 1 m/s^2 downward b) 4 m/s and 2 m/s^2 downward c) 4 m/s and 2 m/s^2 upward d) 6 m/s and 4 m/s^2 downward e) 6 m/s and 4 m/s^2 upward 2. Relevant equations W = ΔKE V(t)^2 = V0^2 + 2 a Δs 3. The attempt at a solution W = ΔKE ma s = (1/2) m v^2 a (2) = (1/2) (4)^2 a = 4 m/s^2 So, the acceleration of block A is 4 m/s^2 downward But, what is the block B acceleration and its velocity when block A goes 2m down ? I need to find the relationship between block A acceleration and block B acceleration. Their accelerations are same in magnitude, right ? (Since the rope is free) So, block B acceleration is 4 m/s^2 downward, right ? But, what about block B velocity ? Using the equation V(t)^2 = V0^2 + 2 a Δs V(t) = √(2aΔs) = √16 = 4 m/s But, my answer is not in the choices.. Please help me..