Question

The system starts at rest. If you pull down the cord at A such that the acceleration of A is a constant 3 mm/sec/sec for 10 cm, what is the resulting change in position, velocity and acceleration of block B? Assume that the pulleys are frictionless, massless and have radius 5 cm.

Answer 1

Acceleration of B will be same as A but in opposite direction is when we pull down and A, , means B will be pulled up: So Acceleration of B = a = 3 mm/s^2 Because u = 0, s = 10 cm = 0.1 m a = 3 m/s^2 = 3 times 10^- 3 m/ s Using, v^2 = u^2 + 2 a s v^2 = 2 times 3 times 10^- 3 times 0.1 = 6 times 10^- 4 Rightarrow v = 2.45 times 10^- 2 m/s upward For position, x = 10 cm Rightarrow x = 0.1 m upward from initial position Acceleration a = 3 mm./s^2 a = 3 times m/s^2 upward