# Three bolocks are connected(mass_1 on the left connected to mass_2 with a string which is T_1. Mass_2 is connected to Mass_3 with a string, T_2. the force is holding on to T_3 which is connected to mass_3) and pulled to the right on a horizontal frictionless table by a force witha magnitude of T_3=65N. If m_1= 12 kg, m_2 =24 kg and m_3= 31 kg, what is a) acceleration of the sysem and b) tensions T_1 AND C) T_2 in the inerconnecting cords.

what is the formula I would use to find a? Is it a= F/m? for the tensions how would I find them?

## To find the acceleration of the system, you are correct in using the formula a = F/m. However, in this case, the force applied is the tension in the string T_3 that is pulling the entire system to the right. So, you will use the equation:

a = T_3 / (m_1 + m_2 + m_3)

Substituting the given values:

a = 65N / (12kg + 24kg + 31kg) = 65N / 67kg ≈ 0.97 m/s²

For finding the tensions T_1 and T_2, we can consider each block separately and apply Newton's second law of motion to each of them.

For mass_1, the equation would be:

T_1 - friction = m_1 * a

As the table is frictionless, the friction term can be ignored. Substituting the given values for m_1 and a:

T_1 = m_1 * a = 12kg * 0.97m/s² = 11.64N

So, the tension T_1 in the string connecting mass_1 and mass_2 is approximately 11.64N.

For mass_2, the equation would be:

T_2 - T_1 = m_2 * a

Substituting the given values for T_1, m_2, and a:

T_2 = T_1 + m_2 * a = 11.64N + 24kg * 0.97m/s² ≈ 35.88N

So, the tension T_2 in the string connecting mass_2 and mass_3 is approximately 35.88N.

To summarize:

a) The acceleration of the system is approximately 0.97 m/s².

b) The tension T_1 is approximately 11.64N, and the tension T_2 is approximately 35.88N.