91Ó°ÊÓ

Speed of the two barges is$v_1=10\mathrm{km}/\mathrm{h},{\mathrm{v}}_2=20\mathrm{km}/\mathrm{h}$

Coal is shoveled from barge 1 to barge 2 with rate$R=1000\mathrm{kg}/\min$

We can find the additional force that must be provided to barge 2 using Newton’s second law which relates force with momentum. From the given information in the problem about frictional force we can guess theadditional force that must be provided by the driving engines of the slower barge (1) if neither is to change speed.

Formula:

$F=\frac{dP}{dt}$

Speed of the two barges is

$$\begin{gathered} v_1=10\mathrm{km}/\mathrm{h} \\ =10×\frac{5}{18} \\ =2.78\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_2=20\mathrm{km}/\mathrm{h}×\frac{5}{18} \\ =5.56\mathrm{m}/\mathrm{s} \\ R=1000\mathrm{kg}/\min \\ =\frac{1000}{60} \\ =16.67\mathrm{kg}/\mathrm{s} \end{gathered}$$

To maintain the constant speed after shoveling the coal from the slower barge 1 to the faster barge 2, the additional force should be provided to barge 2.

According to Newton’s second law,

$$\begin{gathered} F=\frac{dP}{dt} \\ F=\frac{d\left(mv\right)}{dt} \\ F=\frac{d\left(mv\right)}{dt} \\ F=\frac{dm}{dt}\left(v_2-v_1\right) \\ =16.67\left(5.56-2.78\right) \\ =46\mathrm{N} \end{gathered}$$

Therefore, the additional force that must be provided by the driving engines of the faster barge (2) if neither is to change speed is 46 N

The coal is shoveled in y direction from barge 1, and there is no frictional force acting on it in x direction.

Therefore, to maintain a constant speed, the slower barge does not require any additional force.