# Is the net force on centre of mass always equal to the external force applied on one block of a two block system connected by a spring?

I came across a question which says- "Two blocks of equal mass m are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force F is applied on one of the blocks pulling it away from the other as shown in figure. Find the position of centre of mass at time t"

In the solution the acceleration of the centre of mass has been found out by dividing F by total mass of the system i.e.

a(com)=(F/2m) , com: Centre of mass

I don't get how the force applied on one of the blocks is considered the same on the centre of mass. Since the spring offers a resistive force the centre of mass should have moved with a net acceleration a which can be given by the equation-

F- Spring force(kx)=2ma, where F is the external force applied on one block, k is the spring constant, x is the elongation of the spring, 2m is net the mass of system, m being mass of one block and a ia the acceleration of the centre of mass.

The force of the spring, does not act on the center of mass. That is because it does pull one of your masses in one direction, but the other one in the other, so the center of your mass is not affected by it.

Looking only at the force caused by the spring ($$F_S$$) you can see $$F_{S, m_1} = F_{S, m_2} \Leftrightarrow F_{S, m_1} + F_{S, m_2}= 0 \Leftrightarrow A_{S, m_1} m_1 + A_{S, m_2} m_2 = \frac{A_{S, m_1} m_1 + A_{S, m_2} m_2}{m_1 + m_2} \cdot (m_1 + m_2) =A_{S, com} m_{S, com} = F_{S, com} = 0$$.

If now looking at the external force acting on $$m_2$$ only one can see for the center of mass: $$F_{com} = \frac{A_{m_1} m_1 + A_{m_2} m_2}{m_1 + m_2} \cdot (m_1 + m_2) = A_{m_1} m_1 + A_{m_2} m_2 = F_1 + F_2 = F_2 = F_{ext}$$

Adding the 2 contributions together (superposition principle) you get indeed get $$F_{com} = F_{ext}$$.

Now looking at your proposal for the expression for the acceleration due to the restive force of the spring. This does look like the acceleration of the spring, I would expect to act on $$m_2$$, expect that you wrote double the mass there $$F_{S, m_2}(x(t),k) = k \cdot x(t) = m A_{S, m_2}$$. However the resistive force of the spring only acts on mass 2, while it acts with an equally sized accelerating force on $$m_1$$, which is why the contributions to the center of mass cancel out.

For a less mathematical and more intuitive understanding, try to imagine the edge cases of $$k \rightarrow 0$$ (no spring) and $$k \rightarrow \infty$$ (rigid connection between the masses) in both cases it might be seen intuitively (maybe?), that the external force is the same force acting on the center of mass. Then why would one expect anything else in the intermediate cases?

If you do a force Balance on the mass on the right, you get $$F-k(x_R-x_L-d_0)=m\frac{d^2x_r}{dt^2}$$ where x_R and x_L are the coordinates of the masses on the right and the mass on the left, respectively, d_0 is the initial length of the spring, and is the spring constant. For the mass on the left, the force balance is $$k(x_R-x_L-d_0)=\frac{d^2x_L}{dt^2}$$If we add these two force balance together, we get $$F=m\left(\frac{d^2 x_L}{dt^2}+\frac{d^2 x_R}{dt^2}\right)=(2m)a_C$$ where 2m is the total combined mass of the blocks and $$a_C$$ is the acceleration of the center of mass: $$a_C=\frac{d^2}{dt^2}\left(\frac{x_L+x_R}{2}\right)$$

The net force on the centre of mass equals the external force applied on the total system, so in this case the external force applied on both blocks added together, which means just the force on one block because there is none on the other.

That's true for this example, but not in general, so you cannot state it as a rule as you do in the title.