91影视

$$\begin{gathered} {\mathrm{a}}_1=12.0\mathrm{m}/{\mathrm{s}}^2 \\ {\mathrm{a}}_2=3.30\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

This problem deals with the relationship between the three quantities: force, mass and acceleration in Newton鈥檚 second law F = ma.

The 鈥渃ertain force,鈥� denoted as F, is assumed to be the net force on the object when it gives $m_1$an acceleration ${\mathrm{a}}_1=12\mathrm{m}/{\mathrm{s}}^2$and when it gives ${\mathrm{m}}_2$an acceleration $a_2=3.3m/s^2,\mathrm{i}.\mathrm{e}.,\mathrm{F}={\mathrm{m}}_1{\mathrm{a}}_1={\mathrm{m}}_2{\mathrm{a}}_2$.The accelerations for $m_1+m_2\mathrm{and}{\mathrm{m}}_1-{\mathrm{m}}_2$can be solved by substituting $m_1=\frac{F}{a_1}\mathrm{and}{m}_2=\frac{F}{a_2}$.

Now we seek the acceleration aof an object of mass $m_2-m_1$ when F is the net force on it. The result is

role="math" localid="1660907341257" $$\begin{gathered} a=\frac{F}{m_2-m_1} \\ =\frac{F}{\left(F/a_2\right)-\left(F/a_1\right)} \\ =\frac{a_1{a}_2}{a_1-a_2} \\ =\frac{\left(12.0\mathrm{m}/{\mathrm{s}}^2\right)\left(3.30\mathrm{m}/{\mathrm{s}}^2\right)}{12.0\mathrm{m}/{\mathrm{s}}^2-3.30\mathrm{m}/{\mathrm{s}}^2} \\ =4.6\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the solution is$a=4.6\mathrm{m}/{\mathrm{s}}^2$

Similarly for an object of mass $m_2+m_1$, we have:

$$\begin{gathered} a=\frac{F}{m_2+m_1} \\ =\frac{F}{\left(F/a_2\right)-\left(F/a_1\right)} \\ =\frac{a_1{a}_2}{a_1-a_2} \\ =\frac{\left(12.0\mathrm{m}/{\mathrm{s}}^2\right)\left(3.30\mathrm{m}/{\mathrm{s}}^2\right)}{12.0\mathrm{m}/{\mathrm{s}}^2-3.30\mathrm{m}/{\mathrm{s}}^2} \\ =2.6\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Hence},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{a}=2.6\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$