Answer :
Missing part of the text:
"Two masses, m1 = 2.12 kg and m2 = 9.01 kg are on a horizontal frictionless surface and they are connected together with a rope as shown in the figure."
and missing figure (see attachment)
Solution:
We can write Newton's second law for the whole system m1-m2 and for m2 only (2 equations). Only one force (F) acts on the m1-m2 system, while if we consider m2 only we have two forces acting on it: F and T (tension), in the opposite direction. So, the two equations are
[tex]F=(m_1+m_2)a[/tex]
[tex]m_2 a=F-T[/tex]
where a is the acceleration of the system.
From the first equation we get
[tex]a= \frac{F}{m_1+m_2} [/tex]
and substituting it inside the second equation, we get
[tex] \frac{m_2}{m_1+m_2}F=F-T [/tex]
re-arranging, we get
[tex]F=T( \frac{m_1+m_2}{m_1} )[/tex]
Using [tex]m_1=2.12 Kg[/tex], [tex]m_2=9.01 Kg[/tex], and using the maximum value of T that is allowed not to break the rope (T=55 N), we can find the maximum allowed value for F:
[tex]F=288.8 N[/tex]
"Two masses, m1 = 2.12 kg and m2 = 9.01 kg are on a horizontal frictionless surface and they are connected together with a rope as shown in the figure."
and missing figure (see attachment)
Solution:
We can write Newton's second law for the whole system m1-m2 and for m2 only (2 equations). Only one force (F) acts on the m1-m2 system, while if we consider m2 only we have two forces acting on it: F and T (tension), in the opposite direction. So, the two equations are
[tex]F=(m_1+m_2)a[/tex]
[tex]m_2 a=F-T[/tex]
where a is the acceleration of the system.
From the first equation we get
[tex]a= \frac{F}{m_1+m_2} [/tex]
and substituting it inside the second equation, we get
[tex] \frac{m_2}{m_1+m_2}F=F-T [/tex]
re-arranging, we get
[tex]F=T( \frac{m_1+m_2}{m_1} )[/tex]
Using [tex]m_1=2.12 Kg[/tex], [tex]m_2=9.01 Kg[/tex], and using the maximum value of T that is allowed not to break the rope (T=55 N), we can find the maximum allowed value for F:
[tex]F=288.8 N[/tex]
