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Problem8.44 parts A and B

Problem8.44 parts A and B
Problem8.44 parts A and B
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Answer #1

block on the top of table is having twice inertia as compared to first block

since inertia is directly proportional to mass so,

mass of block on top of table will be double of another mass

frictional force = 0.5 * mass of hanging block * g

force equation of block on table

T - frictional force = 2 * mass of hanging block * a

T - 0.5 * mass of hanging block * g = 2 * mass of hanging block * a

force equation of hanging block

mass of hanging block * g - T = mass of hanging block * a

on solving we'll get

a = g / 6

a = 9.8 / 6

acceleration a = 1.633 m/s^2

when block above the table is pushed to the right then frictional force will be in the direction of tension force as friction force is always in opposite direction of motion

so,

equation of motion of block on top of a table

T + 0.5 * mass of hanging block * g = 2 * mass of hanging block * a

equation of motion of hanging block

mass of hanging block * g - T = mass of hanging block * a

on solving we'll get

a = g / 2

acceleration = 4.9 m/s^2

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