Question

Block A and block B are connected by a massless string over a smooth and massless pulley as shown below. The masses of block A and block B are 2 kg and 1 kg respectively The two blocks are released from rest. Ceiling Block B Block A (a) Draw two force diagrams, one for each block. (2%) (b) Calculate the acceleration of block A. (490) (c) Calculate the tension in the rope. (2%) (d) Using the conservation of energy, calculate the velocity of each block after block A has moved down a distance of 0.5 m. (6%). (e) Suppose the whole setup is attached to the ceiling of a lift. The two blocks are initially at rest. The string is taut initially. The cable of the lift suddenly breaks and the whole lift undergoes free-falling motion. What is the value of the tension in the rope in this case? (4%) Cable breaks Block B Block A Lift

Answer 1

Block B T - M = M Block A M_a g - T = M (b) Block A acceleration, g = 10 m/s^2 M g - T = M_a rightdoublearrow 2 times 10 - T = 2a T = 20 - 2a for block B rightarrow T - M_b g = M_b a rightdoublearrow 20 - 2a - 1(10) = a 10 = 3a rightdoublearrow a = 10/3 m/s^2 Acceleration of A. rightdoublearrow a = 3.333 m/s^2 is taking g = 9.8 m/s^2then a = 3.267 m/s^2 a = 3.267 m/s^2 (c) Tension in the rope put value of a in eq^4 (1) T = 20 - a = 20 - 2(3.33) = 20 - 6.66 = 13.34 (d) when A moved A moved down 0.5 m time it takes s = ut - 1/2 at^2 rightdoublearrow 0.5 = 0 times t + 1/2 (3.33) t^2 1/3.33 = t^2 = .003 t = squareroot = 0.548 velocity od A rightarrow