A 8.90-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 0.165, find the tension in the string. (The block slides to the right in the diagram below.)

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Asked Oct 28, 2019
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A 8.90-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 0.165, find the tension in the string. (The block slides to the right in the diagram below.)

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Expert Answer

Step 1

The net horizontal force acting on the object mass m1 is,

The net vertical force acting on the object mass m2 is,

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т-л-та т-та+f 3 та+дтв т,g -Т%3D та

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Step 2

Substitute the tension force acting on the object mass m1 in above relation, to find the acceleration of the block.

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тg - та-дтg- та т,в — дтв-та+та 8(т, - дт)-а(т + т,) 8(т, - дт) т+т,

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Step 3

Substitute the ...

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(9.8m/s')(8.90kg-(0.16s)(5.0kg)) 5.0kg +8.90kg =5.69m/s2

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Science

Physics

Newtons Laws of Motion

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