Mass 1 (10 kg) rests on a frictionless table connected by a string to Mass 2 (5 kg). Find

a) the acceleration of each block. (Draw the figure and proceed).

b) the tension in the connecting string.

To find the acceleration of each block in this setup, we can apply Newton's second law of motion. Let's assume that Mass 1 is on the left and Mass 2 is on the right.

a) To determine the acceleration of each block, we need to consider the net force acting on each block.

For Mass 1:
Since the table is frictionless, the only force acting on Mass 1 is the tension force in the string. Using Newton's second law (F = ma), we can set up the equation:
Tension force = mass * acceleration for Mass 1

For Mass 2:
There are two forces acting on Mass 2: the tension force in the string pulling it to the right, and the tension force on the other side of the string pulling it to the left. Using Newton's second law, we can set up the equation:
Tension force pulling to the right - Tension force pulling to the left = mass * acceleration for Mass 2

b) To find the tension in the connecting string, we can use the equation derived for Mass 1 in part a).

Here's an example diagram of the setup:

-------------- <- Table
| |
o-----------T
|-> Mass 1 (10 kg)
|
|
-------T---------o <- Mass 2 (5 kg)

Now, let's proceed with solving the equations:

For Mass 1:
Tension force = mass_1 * acceleration_1

For Mass 2:
(Tension force pulling to the right) - (Tension force pulling to the left) = mass_2 * acceleration_2

Now, using the concept that the tension in the connecting string is the same for both masses (since it is a single string), we can equate the two equations:

mass_1 * acceleration_1 = (Tension force pulling to the right) - (Tension force pulling to the left)

Given that mass_1 is 10 kg and mass_2 is 5 kg, we can substitute these values into the equations.

Substituting mass_1 = 10 kg and mass_2 = 5 kg into the equations, we get:

10 * acceleration_1 = (Tension force pulling to the right) - (Tension force pulling to the left)

Substituting acceleration_1 = acceleration_2 = a (since the same string connects them), we have:

10a = (Tension force pulling to the right) - (Tension force pulling to the left)

Now, let's solve for the tensions:

The tension force pulling to the right is the tension in the string connecting Mass 1 to Mass 2. Let's call this T_1.

The tension force pulling to the left is the tension in the string connecting Mass 2 to the table. Let's call this T_2.

Therefore, we have:

10a = T_1 - T_2 (equation 1)

For Mass 2, we have:

(T_1) - (T_2) = 5a (equation 2)

Now, we have a system of equations that we can solve simultaneously. Subtract equation 2 from equation 1:

10a - (T_1 - T_2) = T_1 - T_2 - 5a

Simplifying:

15a = 2T_1 - 2T_2

Now, to find the tension in the connecting string (T_1), we need to eliminate T_2. We can do this by assuming that the tension on the other side of the string (T_2) is 0 (since the table is frictionless).

This simplifies the equation to:

15a = 2T_1

Now we can solve for the acceleration and tension:

a = (2/15) T_1

Therefore, the acceleration of each block is (2/15) times the tension in the connecting string. To find the tension in the string, we would need additional information about the system, like the applied force, any given acceleration, or angle of the incline, etc.