A block of mass m1 = 4.4 kg rests on a frictionless horizontal surface. A second

Question

A block of mass m1 = 4.4 kg rests on a frictionless horizontal surface. A second block of mass m2 = 1.1 kg hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block. The blocks are released from rest. (a) Find the acceleration of the two blocks after they are released. block 1: magnitude Incorrect: Right block 2: m/s2 direction downward (b) What is the velocity of the first block 1.8 s after the release of the blocks, assuming the first block does not run out of room on the table and the second block does not land on the floor? magnitude m/s direction: to the right. (c) How far has block 1 moved during the 1.8-s interval? ________m (d) What is the displacement of the blocks from their initial positions 0.45 s after they are released? block 1: magnitude m direction: to the right block 2: magnitude m direction: down

Explanation / Answer

let T be the tension on the string and "a" be the acceleration of both block after released let the acceleration of M2 in downward direction is "a " ,, then the M1 block will also move with acceleration "a " now let tension on the string is T ,,, balancing force for M2 ,, downword force is M2* g and upward force is T ,, => M2*g- T= M2*a : ...1 balaccing force for M1 => T = M1*a .......2 putting the value of T from 2 nd equation in 1 st equation => M2*g - M1*a = M2*a => M2*g = M2*a +M1*a =>a= M2*g/ M1+M2 m/s^2 =1.96 m/s^2 velocity of first block and second block after 1.8 s = > v= u +a*t => v= 0 + (M2g/M1+M2 ) *1.8 = M2g/M1+M2 *1.8m/s = 3.528 m/s distance moved by first block in 1.8 s = > d= u*t + a*t^2 /2 = 0+ 1.96*1.8^2*1/2= 3.1752 meter displacement of 1st and second block after .45 s = > d = u*t + a*t^2 /2 = 0+ 1.96*.45^2*1/2 = 0.19845

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