Answer:
The correct answer is "12 m/s²".
Explanation:
Given:
[tex]F_{app} = 100 \ N[/tex]
As we know,
⇒ [tex]F_{app} = mg-ma[/tex]
Or,
⇒ [tex]a = g-(\frac{F_{app}}{m} )[/tex]
By substituting the values, we get
⇒ [tex]=10-(-\frac{100}{50} )[/tex]
⇒ [tex]=10+2[/tex]
⇒ [tex]=12 \ m/s^2[/tex]
