Consider the situation shown in figure (18-E16). The elevator is going up with an acceleration of 2.00 ms-2 and the focal length of the mirror is 12.) cm. All the surfaces are smooth and the pulley is light. The mass-pulley system is released from rest (with respect to the elevator) at t=0 when the distance of B from the mirror is 42.) cm. Find the distance between the image of the block B and the mirror at t=0.200 s. Take g= 10 ms-2


acceleration of the elevator, a= 2.00 ms-2

focal length of the mirror M, f= 12 cm


mass of the blocks A and B = m


mass pulley system is released at time t=0


let the acceleration of the masses A and B w.r.t. the elevator = a



using free body diagram




using both the equations,




distance travelled by the block B of mass m in time t= 0.2 sec is given by





It is mentioned in the question that the distance of block B from the mirror is 42 cm.


The object distance u from the mirror = -(42-12) = -30cm


By using mirror formula,





Distance between image of block B and mirror = 8.57 cm.


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