A block A of mass 8 kg is placed on a frictionless horizontal table. A thread tied to it passes over a frictionless pulley and carries a body B of mass 2 kg at the other end. Find the acceleration of the system. Also find the tension in the thread. If the thread is cut into two and tied to the ends of a spring of forceconstant 1600N/m, find the amount of stretching of the spring. Neglect the mass of thread and of spring `(g = 9.8 m//s^(2))`

Explanation

Figure(a) represents the arrangement of the first part of the problem
Here `m_1 = 8kg , m_2 = 2kg `
If a is acceleration of system and T the tension in string

then we have for the motion of block B ,
`m_2 g - T = m_2 a ` ....... (1)
and for the motion of block A , T = ` m_1 a ` ....... (2)
( since weight `m_1g` is balanced by normal reaction R )
Adding (1) and (2) , we get ` m_2 g = (m_1 + m_2) a ` `:. ` acceleration ` a=(m_2)/(m_1 + m_2) g` ...... (3)
Substituting this is (2) , we get , Tension `T=(m_1 m_2)/( m_1 + m_2 ) g `.......... (4)
In second part the spring is introduced and the arrangement is shown in figure (b) .
The spring is pulle by Tension `T=(m_1 m_2)/( m_1 + m_2) g `
If x is the streching of spring , then we have
`T= Kx (or ) x =(T)/(K) = (m_1 m_2 g)/( (m_1 + m_2)K)`
substituting ` m_1 = 8 kg , m_2 = 2kg ` .
K= 1600 N/m in (3), (4) and (5) , we get acceleration
` a= ( m_2)/( m_1 +m_2)g= (2) /( 2 + 8 ) xx 9.8 = 1.96 m//s^(2)`
Tension `T= (m_1 m_2)/( m_1 + m_2)g = (2 xx8)/(2 + 8) xx 9.8 = 15.68` N
and stretching ` x= (T)/(K) = (15.68)/(1600) = 0.0098 m` .