In a trolley and block system, a block of mass \( 3\text{ kg} \) hangs vertically and is connected by a light inextensible string passing over an ideal pulley to a trolley of mass \( 20\text{ kg} \) moving on a horizontal track. The coefficient of kinetic friction between the trolley and the track is \( \mu_k = 0.04 \). Calculate the tension \( T \) in the connecting string. (Take \( g = 10\text{ m s}^{-2} \))

Let the mass of the hanging block be \( m = 3\text{ kg} \), and the mass of the trolley be \( M = 20\text{ kg} \). \nFor the hanging block moving downwards with acceleration \( a \): \n\( m g - T = m a \implies 3(10) - T = 3a \implies 30 - T = 3a \)\nFor the trolley moving rightwards on the horizontal track with the same acceleration \( a \): \n\( T - \mu_k M g = M a \implies T - 0.04(20)(10) = 20a \implies T - 8 = 20a \)\nAdding both equations to find \( a \): \n\( (30 - T) + (T - 8) = 3a + 20a \)\n\( 22 = 23a \implies a = \frac{22}{23}\text{ m s}^{-2} \approx 0.9565\text{ m s}^{-2} \)\nNow, substitute \( a \) back into the equation for tension \( T \): \n\( T = 30 - 3a = 30 - 3(\frac{22}{23}) = 30 - \frac{66}{23} = 30 - 2.87 = 27.13\text{ N} \approx 27.1\text{ N} \).