Question 5.7: One Block Pushes Another Two blocks of masses m1 and m2, wit...

(A) Conceptualize Conceptualize the situation by using Figure 5.13a and realize that both blocks must experience the same acceleration because they are in contact with each other and remain in contact throughout the motion.

Categorize We categorize this problem as one involving a particle under a net force because a force is applied to a system of blocks and we are looking for the acceleration of the system.

Analyze First model the combination of two blocks as a single particle under a net force. Apply Newton’s second law to the combination in the x direction to find the acceleration:

(1)   a_x=\frac{F}{m_1+m_2}

Finalize The acceleration given by Equation (1) is the same as that of a single object of mass m_1 + m_2 and subject to the same force.

(B) Conceptualize The contact force is internal to the system of two blocks. Therefore, we cannot find this force by modeling the whole system (the two blocks) as a single particle.

Categorize Now consider each of the two blocks individually by categorizing each as a particle under a net force.

Analyze We construct a diagram of forces acting on the object for each block as shown in Figures 5.13b and 5.13c, where the contact force is denoted by \overrightarrow{P}. From Figure 5.13c, we see that the only horizontal force acting on m_2 is the contact force \overrightarrow{P}_{12} (the force exerted by m_1 on m_2 ), which is directed to the right.

Apply Newton’s second law to m_2 :

(2) \sum F_x=P_{12}=m_2 a_x

Substitute the value of the acceleration a_x given by Equation (1) into Equation (2):

(3) \quad P_{12}=m_2 a_x=\left(\frac{m_2}{m_1+m_2}\right) F

Finalize This result shows that the contact force P_{12} is less than the applied force F. The force required to accelerate block 2 alone must be less than the force required to produce the same acceleration for the two-block system.

To finalize further, let us check this expression for P_{12} by considering the forces acting on m_1, shown in Figure 5.13b. The horizontal forces acting on m_1 are the applied force \overrightarrow{F} to the right and the contact force \overrightarrow{P}_{21} to the left (the force exerted by m_2 on m_1 ). From Newton’s third law, \overrightarrow{P}_{21} is the reaction force to \overrightarrow{P}_{12}, so P_{21}=P_{12}.

Apply Newton’s second law to m_1 :

(4) \sum F_x=F-P_{21}=F-P_{12}=m_1 a_x

Solve for P_{12} and substitute the value of a_x from Equation (1):

This result agrees with Equation (3), as it must.