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 . Alsofind 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)).

A block A of mass 8 kg is placed on a frictionless horizontal table .

1.	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 . Alsofind 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)).
Answer» Solution :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 byTension `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`