Calculate the Acceleration of a 4.0 kg Block on a Table

We can solve this problem using Newton's second law. Let's consider the forces acting on the 4.0 kg block.

There are two forces acting on the block: the weight (mg) and the tension in the string (T). Since the block is sliding on the table, we know that the normal force acting on the block must be equal and opposite to the weight. Therefore, the net force acting on the block is given by:

Fnet = T - mg

According to Newton's second law, the acceleration of the block is given by:

a = Fnet / m

Substituting the values, we get:

a = (T - mg) / m

Now, let's consider the forces acting on the 2.0 kg block.

There are two forces acting on the block: the tension in the string (T) and the weight (mg). Since the block is hanging vertically, we know that the tension in the string must be equal and opposite to the weight. Therefore, the net force acting on the block is zero.

Since the net force acting on the system is the tension in the string, we can write:

Fnet = T

And since the total mass of the system is 4.0 kg + 2.0 kg = 6.0 kg, we can write:

a = Fnet / mtotal

Substituting the values, we get:

a = T / mtotal

Since the system is released from rest, we know that the tension in the string will be equal to the weight of the 2.0 kg block. Therefore:

T = mg = (2.0 kg) (9.8 m/s^2) = 19.6 N

Substituting this value into the equation for the acceleration of the 4.0 kg block, we get:

a = (T - mg) / m = (19.6 N - (4.0 kg) (9.8 m/s^2)) / 4.0 kg = 1.5 m/s^2

Therefore, the acceleration of the 4.0 kg block sliding on the table is most nearly 1.5 m/s^2.