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Two blocks of masses 10 kg and 5 kg are connected by a flexible but inextensible string over a shaft as shown in the figure. The system starts from rest with 10 kg mass moving downward. If a constant frictional force of 25 N acts at the shaft, the velocity when the 10 kg block has moved 2 m down is

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$6.2 {ms}^{- 1}$

$2.6 {ms}^{- 1}$

$3.2 {ms}^{- 1}$

$4 {ms}^{- 1}$

answer is A.

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Detailed Solution

$a = \frac{10 g - 5 g - 25}{15} = \frac{100 - 50 - 25}{15} = \frac{5}{3} {ms}^{- 2}$

$Since, v^{2} - 0^{2} = 2 as$

$\begin{array}{l} \Rightarrow v^{2} = 2 (\frac{5}{3}) (2) = \frac{20}{3} \\ \Rightarrow v = 2 .6 {ms}^{- 1} \end{array}$

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