Two blocks, joined by a string, have masses of 6 and 9kg. They rest on frictionless horizontal surface. A 2nd string, attached only to the 9kg block, has a horizontal force = 30N applied to it. Both blocks accelerate. Find the tension in the string between the blocks.

Both blocks accelerate at the same rate. Draw free body diagrams and write Newton's second law (F = ma) for each block. Solve for the two unknowns: accleration and string tension.

To find the tension in the string between the blocks, we can start by analyzing the forces acting on each block and using Newton's second law of motion (F = ma) to determine their accelerations.

Let's consider the smaller block with a mass of 6 kg:
- The only force acting on it is the tension force, which we'll call T1.
- The acceleration of this block will be the same as the system's acceleration since it's connected to the other block through the string.

Applying Newton's second law to this block, we have:
T1 = m1 * a

Next, let's consider the larger block with a mass of 9 kg:
- It has two forces acting on it: the tension force T1 from the string and the applied horizontal force of 30 N.
- The tension force T1 acts in the direction of motion and the applied force counteracts it.
- The acceleration of this block will be the same as the system's acceleration but in the opposite direction due to the applied force.

Applying Newton's second law to this block, we have:
T1 - 30 N = m2 * (-a) (negative sign in front of 'a' because the acceleration is in the opposite direction)

Now, we need to eliminate the acceleration 'a' from the two equations. To do that, we can substitute the second equation into the first equation to eliminate 'a' and solve for T1.

Substituting the second equation into the first equation, we get:
T1 = m1 * (- T1/m2 - 30 N)

Simplifying the equation, we have:
T1 = - (m1 * T1) / m2 - 30 N

To solve for T1, we can rearrange the equation:
T1 + (m1 * T1) / m2 = - 30 N

Combining like terms:
(m2 * T1 + m1 * T1) / m2 = - 30 N

Factoring out T1:
T1 * (m2 + m1) / m2 = - 30 N

Now, we can isolate T1 by multiplying both sides by m2:
T1 * (m2 + m1) = - 30 N * m2

Finally, solving for T1:
T1 = (-30 N * m2) / (m2 + m1)

Now, substitute the given values:
m1 = 6 kg
m2 = 9 kg
Force = 30 N

T1 = (-30 N * 9 kg) / (9 kg + 6 kg)
T1 = -270 N / 15 kg
T1 = -18 N

Since tension cannot be negative, the magnitude of the tension in the string between the blocks is 18 N.