Answer :
Answer:[tex]m_1g<T<m_2g[/tex]
Explanation:
Given
Two masses [tex]m_1[/tex] and [tex]m_2[/tex] is released and there is tension T in the string
Suppose a is the acceleration of the system
Therefore from Diagram
For [tex]m_1[/tex]
[tex]T-m_1g=m_1a[/tex]
[tex]T=m_1(g+a)[/tex]------1
for m_2 body
[tex]m_2g-T=m_2a[/tex]
[tex]T=m_2(g-a)[/tex]-------2
From above two Equation it is said that Tension is greater than m_1g and less than m_2g
[tex]m_1g<T<m_2g[/tex]
Answer:
T<m2hh
Explanation:
Two objects having masses m1 and m2 are connected to each other as shown in the figure and are released from rest. There is no friction on the table surface or in the pulley. The masses of the pulley and the string connecting the objects are completely negligible. What must be true about the tension t in the string just after the objects are released?
Two masses m1 and m2 is released and there is tension T in the string
a is the acceleration of the system
Therefore
For m1
T-m1g=m1a
T=m1(g+a)
m2
this is so because te weight is going downward wile tension in te rope is facing upward
m2g-T=m2a
T=m2(g-a)
from the equation ,
T<m2