Ncert Exercises & Solutions

Q. 28 Two masses of 5 kg and 3 kg are suspended with help of massless inextensible strings as shown in figure. Calculate $T_{1} and T_{2}$ $T_{1} and T_{2}$ when the whole system is going upwards with acceleration $= 2 m / s^{2} (use g = 9 .8 {ms}^{- 2})$ $= 2 m / s^{2} (use g = 9 .8 {ms}^{- 2})$ .

Hint: Apply Newton's second law of motion

Step 1: Apply Newton's second law on both of the bocks separately.

$\begin{array}{l} Given, m_{1} = 5 kg, m_{2} = 3 kg \\ g = 9 .8 m / s^{2} and a = 2 m / s^{2} \end{array}$

$\begin{array}{l} For the upper block \\ \begin{matrix} T_{1} - T_{2} - 5 g = 5 a \\ \Rightarrow T_{1} - T_{2} = 5 (g + a) . . . . . . . . . . . . . . . . . (1) \\ For the lower block T_{2} - 3 g = 3 a . . . . . . . . . . . . . (2) \\ Step 2 : Solve the above equations \\ \Rightarrow T_{2} = 3 (g + a) = 3 (9 .8 + 2) = 35 .4 N \\ From Eq. (i) T_{1} = T_{2} + 5 (g + a) \\ = 35 .4 + 5 (9 .8 + 2) = 94 .4 N \end{matrix} \end{array}$

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