question_answer 68)

                  Two masses of 5 kg and 3 kg are suspended with help of massless inextensible strings as shown in Fig. Calculate \[{{T}_{1}}\]  and \[{{T}_{2}}\]  when whole system is going upwards with acceleration \[=2\,m{{s}^{-2}}\] (use \[g=9.8\,m\,{{s}^{-2}}\]).

Answer:

                  Motion of block of mass 5 kg (say \[{{m}_{1}}\])                 \[{{m}_{1}}a={{T}_{1}}-({{T}_{2}}+{{m}_{1}}\,g)\]                 \ \[{{T}_{1}}{{T}_{2}}={{m}_{1}}\,(a+g)=5\,(11.8)=59\]          ?(i)                 For the motion of block of 3 kg                 \[{{m}_{2}}a={{T}_{2}}{{m}_{2}}g\]                 or \[{{T}_{2}}={{m}_{2}}(a+g)=3(2+9.8)\]                 \[=35.4\,N\]                 From eqn. (i), \[{{T}_{1}}=59+{{T}_{2}}=59+35.4\]             \[=\text{ }94.4N.\]