Consider the situation shown in the figure. The elevator is travelling upwards with an acceleration of 2.00 m s ⁻² and the focal length of the mirror is 12.0 cm . All the surfaces are smooth and the pulley is light. At t = 0 , the mass–pulley system is released from rest (with respect to the elevator) when the distance of block B from the mirror is 42.0 cm . Determine the distance between the image of the block B and

Step-by-step solution

In the frame of the elevator, the effective acceleration due to gravity is: g_ eff = g + a = 10 + 2.00 = 12.0 m s ⁻² Let a_ rel be the acceleration of the blocks relative to the elevator. Writing the equations of motion for blocks A and B: For block A: T = m a_ rel For block B: m g_ eff - T = m a_ rel Adding the two equations, we get: m g_ eff = 2m a_ rel a_ rel = g_ eff 2 = 12.0 2 = 6.00 m s ⁻² The distance fallen by block B relative to the elevator in time t = 0.200 s is: s = 1 2 a_ rel t^2 = 1 2 (6.00) (0.200)^2