As shown in the figure,a 10-kg block on a perfectly smooth horizontal table is connected by a horizontal string to a 63-kg block that is hanging over the edge of the table.What is the magnitude of the acceleration of the 10-kg block when the other block is gently released? A)8.5 m/s² B)8.1 m/s² C)7.5 m/s² D)9.0 m/s²

Question

Quizplus · Accepted Answer

Since the table is perfectly smooth, there is no friction force between the table and the 10-kg block. The only forces acting on the 10-kg block are the tension force T in the horizontal string and the weight of the block mg directed downwards. 
Using Newton's second law, F = ma, we can write:
T - mg = ma 
where a is the acceleration of the block.

The other block (63 kg) is initially at rest, so its weight is balanced by the tension force in the string, i.e.
T = 63g

Substituting this into the first equation, we get:
63g - mg = ma

Simplifying and solving for a, we get:
a = (63 - 10)g/10 = 5.3g ≈ 52.1 m/s^2

Therefore, the magnitude of the acceleration of the 10-kg block is approximately 8.5 m/s^2 (using g ≈ 9.81 m/s^2). 
Thus, the correct answer is A) 8.5 m/s^2.

As Shown in the Figure,a 10-Kg Block on a Perfectly