Question

Two blocks are in contact on a frictionless table. A
horizontal force is applied to the larger block, as shown in the figure.
If m1 = 2.3 kg, m2 = 1.2 kg, and F = 3.2 N,
a) find the magnitude of the force between the two blocks.
b) Show that if a force of the same magnitude F is applied to the
smaller block but in the opposite direction, the magnitude of the
force between the blocks is 2.1 N, which is not the same value
calculated in (a).
c) Explain the difference.

Answer 1

Answer:

Explanation:

The force = m a

Therefore a=m/f

Answer 2

Given:

m1 = 2.3 kg

m2 = 1.2 kg

F = 3.2 N

(a) $\left|F_{21}\right|=\left|F_{12}\right|$

from m2 block

$\begin{array}{l}F_{12}=m_{2} a \\\\a=\frac{F_{12}}{m_{2}}\end{array}$

then m1 block

$\begin{array}{c}F-F_{21}=m_{1} a \Rightarrow F-F_{21}=m_{1} \frac{F_{12}}{m_{2}} \\\\F_{21}\left[1+\frac{m_{1}}{m_{2}}\right]=F\end{array}$

$F_{21}=\frac{m_{2} F}{m_{1}+m_{2}}$

$=\frac{1.2}{1.2+2.3} \times 3.2$

$F_{21}=1.1 \mathrm{~N}$

(b) situation is as shown in figure then similarly

$F_{21}^{\prime}=\frac{2.3}{2.3+1.2}(3.2)$

$F_{21}^{\prime}=2.1\ N$

we know that in (a) $F_{21}$ is on $m_{2}$  while in (b) $F_{21}$ is on $m_{1}$ ,is only horizontal force

$m_{1}>m_{2}$  and $F_{21}=m_{2} a$ in (a) and $F_{2,}=m_{1} a$ in (b) for them to be same

$F_{21}^{\prime}>F_{21}$

Hence the answer are :

(1) 1.1 N

(2) 2.1 N

(2) Difference $F_{21}^{\prime}>F_{21}$