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Blockk on a table with a pulley

  • Thread starter atwood9
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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
 

Answers and Replies

  • #2
arildno
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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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