Question

free - response
two blocks, of mass m and 2m, respectively, are being pushed along a smooth horizontal surface by a force f0, as shown above.
block of mass m
block of mass 2m
a)the two dots, below, represent each of the blocks shown above. on each dot, draw a free - body diagram of the corresponding block. each force should be represented by a single arrow starting on the dot.
b)consider the two cases below. in case a, the force f0 is exerted on to the left side of the smaller block and directed to the right. in case b, the force f0 is exerted on to the right side of the larger block and directed to the left.
case a
case b
c)in a clear, coherent, paragraph - length response, indicate whether the acceleration of the blocks is greater in case a or case b, or the same in both cases, and explain your reasoning. also indicate whether the force of contact that block m exerts on block 2m is greater in case a, case b or the same in both cases, and explain your reasoning.

Explanation:

Step1: Draw free - body diagrams for part a

For the block of mass $M$: There is the applied force $F_0$ to the right, the normal force $N_M$ upward, and the gravitational force $Mg$ downward. Also, there is a contact force $F_{contact}$ exerted by the block of mass $2M$ to the left.
For the block of mass $2M$: There is the contact force $F_{contact}$ (equal in magnitude and opposite in direction to the contact force on $M$) to the right, the normal force $N_{2M}$ upward, and the gravitational force $2Mg$ downward.

Step2: Calculate acceleration for part c (using Newton's second law $F = ma$)

Treat the two - block system as a single entity. The total mass of the system is $M + 2M=3M$. According to Newton's second law $F = ma$, the acceleration of the system in both cases $a=\frac{F_0}{M + 2M}=\frac{F_0}{3M}$. So the acceleration is the same in both cases A and B.

Step3: Calculate contact force for part c

In case A: Let the contact force between the blocks be $F_{AB}$. For the block of mass $2M$, using Newton's second law $F = ma$, we have $F_{AB}=2Ma$. Substituting $a = \frac{F_0}{3M}$, we get $F_{AB}=\frac{2F_0}{3}$.
In case B: Let the contact force between the blocks be $F_{BA}$. For the block of mass $M$, using Newton's second law $F = ma$, we have $F_{BA}=Ma$. Substituting $a=\frac{F_0}{3M}$, we get $F_{BA}=\frac{F_0}{3}$. So the contact force is greater in case A.

Answer:

a) For the block of mass $M$: Three forces - $F_0$ (right - ward), $N_M$ (up - ward), $Mg$ (down - ward) and $F_{contact}$ (left - ward). For the block of mass $2M$: Three forces - $F_{contact}$ (right - ward), $N_{2M}$ (up - ward), $2Mg$ (down - ward).
c) The acceleration of the blocks is the same in both cases. Reason: Using Newton's second law for the combined system of mass $3M$ with a net force $F_0$, the acceleration $a=\frac{F_0}{3M}$ for both cases. The contact force that block $M$ exerts on block $2M$ is greater in case A. Reason: In case A, for block $2M$, $F_{contact}=2Ma=\frac{2F_0}{3}$, and in case B, for block $M$, $F_{contact}=Ma=\frac{F_0}{3}$.