Determine the time in which the smaller block | vedclass

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(D) The kinetic friction force acting between the blocks is $f_k = \mu N = 0.3 	imes (2 	imes 10) = 6\ N$.For the $2\ kg$ block,the net force is $F_

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$2.19$

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(D) The kinetic friction force acting between the blocks is $f_k = \mu N = 0.3 	imes (2 	imes 10) = 6\ N$.For the $2\ kg$ block,the net force is $F_{net,1} = 10 - 6 = 4\ N$. Its acceleration is $a_1 = \frac{4}{2} = 2\ m s^{-2}$.For the $8\ kg$ block,the only horizontal force is the friction force $f_k = 6\ N$ acting in the forward direction. Its acceleration is $a_2 = \frac{6}{8} = 0.75\ m s^{-2}$.The relative acceleration of the $2\ kg$ block with respect to the $8\ kg$ block is $a_{rel} = a_1 - a_2 = 2 - 0.75 = 1.25\ m s^{-2}$.Using the equation of motion $s = ut + \frac{1}{2} a_{rel} t^2$,where $s = 3.0\ m$ and $u = 0$:$3.0 = 0 + \frac{1}{2} 	imes 1.25 	imes t^2$$t^2 = \frac{3.0 	imes 2}{1.25} = \frac{6}{1.25} = 4.8$$t = \sqrt{4.8} \approx 2.19\ s$.

Determine the time in which the smaller block reaches the other end of the bigger block as shown in the figure. $(g = 10\ m s^{-2})$.

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