Blockk on a table with a pulley

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The discussion focuses on calculating the tension in a string connecting an 8.10-kg hanging object to a 5.00-kg block sliding on a table, with a coefficient of kinetic friction of 0.190. The equations provided include T - F_k = m_1 * a for the block on the table and T - m_2 * g = -m_2 * a for the hanging mass. The common acceleration "a" is crucial for solving the system of equations. The analysis confirms that the forces acting on both blocks must be considered to determine the tension accurately.

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A 8.10-kg hanging object is connected by a light, inextensible cord over a light, frictionless pulley to a 5.00-kg block that is sliding on a flat table. Taking the coefficient of kinetic friction as 0.190, find the tension in the string.

I don't know if my equations are right or where to go from here:
T-m2g=m2a2
Fk-T-m1gsin(90)=m1a1
 
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"Fk-T-m1gsin(90)=m1a1"
What does this mean?

You have TWO forces acting on the horizontally lying block, the tension, which we will say drives the block in the positive x-direction, the friction in the negative.

Furthermore, as block 2 accelerates DOWNWARDS, a direction you with gravity has said is in a negative direction, block 1 is accelerating equally in the positive x-direction. We call that common acceleration "a"

Thus, we write:
T-F_k=m_1*a
T-m_2g=-m_2*a
 

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