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)=8 kg, 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_(1)g` 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 in (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 pulled by Tension `T=(m_(1)m_(2))/(m_(1)+m_(2)) g`
If x is the strtching of spring , then we have
`T=Kx "(or) x=T/K =(m_(1)m_(2)g)/((m_(1)+m_(2))K)`
substituting `m_(1) =8kg , m_(2)=2kg`,
`K=1600N//m `in (3),(4) and (5) , we get acceleration
`a=m_(2)/(m_(1)+m_(2))g=2/(2+8)xx9.8=1.96 m//s^(2)`
Tension `T=(m_(1)m_(2))/(m_(1)+m_(2))g=(2xx8)/(2+8)xx9.8= 15.68N`
and stretching `x=T/K=(15.68)/(1600)=0.0098m`