Answer to Question #168848 in Mechanics | Relativity for April

Question

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

Answer 1

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

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

Equation of motion for block resting on table, $3.5a=T$

Equation of motion for hanging block, $0.5a =0.5*g - T$

where g is acceleration due to gravity.

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

and $T=4.375N$

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