A block of mass m1=6.6 kg rests on a frictionless horizontal surface. A second block of...

A block of mass m1=6.6 kg rests on a frictionless horizontal surface. A second block of mass m2=9.4 kg hangs from an ideal cord of negligible mass, which runs over an ideal pulley and then is connected to the side of the first block. The blocks are released from rest. How far will block 1 move during the 1.1 second interval?

Expert Solution

ANSWER :

GIVEN DATA : Mass (M1) = 6.6 kg ; Mass (M2) = 9.4 kg ; time (t) = 1.1 second

Clue given : 1, Blocks are released from rest ( intial velocity U = 0)

2. mass m1 is on horizontal surface ,m2 is on pulley

So, Block m1 slides horizontally and Block m2 falls vertically.

So the acceleation (a) of the two blocks after they are released is

a = g/[1+(m1/m2)] = 9.8 /[1+(6.6/9.4)] = 5.7575 m/s2

when block is released from rest (U = 0) block 1 moved during the 1.1 second interval is calculated using the kinematic equation

S = U t + (1/2) a t ² = 0 x 1.1 + (1/2) x 5.7575 x 1.1 ² = 3.483 m

Answer is S= 3.483 m

genius_generous answered 2 years ago

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