For the pulley system shown in the figure,cal | vedclass

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(B) The system consists of two masses $m_1 = 4\,kg$ and $m_2 = 6\,kg$ connected by a string over a pulley.The acceleration $a$ of the system is given

Vedclass · Accepted Answer

$300$

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(B) The system consists of two masses $m_1 = 4\,kg$ and $m_2 = 6\,kg$ connected by a string over a pulley.The acceleration $a$ of the system is given by the formula $a = \frac{(m_2 - m_1)g}{m_1 + m_2}$.Substituting the values: $a = \frac{(6 - 4) 	imes 10}{6 + 4} = \frac{2 	imes 10}{10} = 2\,m/s^2$.Using the equation of motion $v = u + at$,where initial velocity $u = 0$:$v = 0 + 2 	imes 5 = 10\,m/s$.The kinetic energy $K$ of the $6\,kg$ block is given by $K = \frac{1}{2}mv^2$.$K = \frac{1}{2} 	imes 6 	imes (10)^2 = 3 	imes 100 = 300\,J$.

For the pulley system shown in the figure,calculate the kinetic energy of the $6\,kg$ block after $5\,s$. (Assume $g = 10\,m/s^2$)

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