Answer :
Answer:[tex]4\ m/s^2[/tex]
Explanation:
Given
mass of toy [tex]m=0.5\ kg[/tex]
Force by first dog [tex]F_1=140\ N[/tex]
Force by second dog is [tex]F_2=138\ N[/tex]
Net force on the dog is
[tex]F_{net}=F_1-F_2=140-138[/tex]
[tex]F_{net}=2\ N[/tex]
and [tex]F_{net}=ma[/tex]
[tex]a=\dfrac{F_{net}}{m}[/tex]
[tex]a=\dfrac{2}{0.5}[/tex]
[tex]a=4\ m/s^2[/tex]
So, horizontal acceleration of the toy is [tex]4\ m/s^2[/tex]