Question #168848

A 3.5-kg block rest on a horizontal tabletop. A cord is attached to this block, the cord passing

over a pulley at the edge of the table. At the end of the cord is hung another block weighing

5 N. Determine the tension in the cord and the distance the blocks moved after 2 seconds


1
Expert's answer
2021-03-04T11:50:51-0500

Let a be the acceleration of the system and T be the tension in the cord.


Mass of the hanging block, M=5/10=0.5KgM = 5/10=0.5Kg


Equation of motion for block resting on table, 3.5a=T3.5a=T


Equation of motion for hanging block, 0.5a=0.5gT0.5a =0.5*g - T

where g is acceleration due to gravity.



Solving these, we get, a=1.25m/s2a=1.25m/s^2

and T=4.375NT=4.375N


Distance covered by block after 2 seconds, d=12at2=0.51.254=2.5md=\frac{1}{2}at^2=0.5*1.25*4=2.5 m


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