In Figure 5-51a, a constant horizontal force ${\overrightarrow{\mathbf{F}}}_{\mathbf{a}}$is applied to block A, which pushes against block Bwith aforce directed horizontally to the right. In Figure 5-51b, the same force${\overrightarrow{\mathbf{F}}}_{\mathbf{B}}$is applied to block B; now block Apushes on block Bwith a 10.0Nforce directed horizontally to the left. The blocks have a combined mass of 1,2 kg. (a) What is the magnitude of their acceleration in Figure 5-52 (a)and (b) What is the magnitude of forceF?

1. Combined mass$12.0kg$
2. Block A pushes against block B with force 20.0N
3. Block B pushes against block A with force 10.0N

Newton’s second law states that the net force on the object is equal to the product of mass and acceleration. Apply Newton’s second law to the system of two blocks and the individual block.

As the applied force isthesame and the mass of both blocks is the same in both diagrams, the acceleration will also be the same.

As block A applies force the force 20.0N is applied to A and when it applies 10.0N force, the force is applied to blockB. Therefore, mass of blockB,

$M_B=2\times M_A$

Now, we have

$$\begin{gathered} M_B+M_A=12.0kg \\ 2M_A+M_A=12.0kg \\ {M}_A=4.0kg \\ {M}_B=8.0kg \end{gathered}$$

Now, according to fig, 5.51a in the above problem

Acceleration is,

$$\begin{gathered} a=\frac{20.0N}{8.0kg} \\ =2.5m/s^2 \end{gathered}$$

The acceleration of the object is$2.5m/s^2$ .

Applying Newton’s second law,

$$\begin{gathered} F_a=12.0kg\times 2.5m/s^2 \\ =30.0N \end{gathered}$$

Therefore, the total force is 30.0N .