Respuesta :
Answer:
[tex]v = \sqrt{\frac{2(m_{b} gd - \mu_{k} m_{a}gd)}{m_{b} +m_{a} + \frac{I}{R^{2} } } }[/tex]
Explanation:
Energy = force * distance
The heat energy lost by the block B equals the sum of the energy gained.
Energy lost = sum of energy gained
Energy lost by [tex]M_{b}[/tex] due to the height covered= KE gained by [tex]M_{b}[/tex] + KE gained by pulley + KE gained by [tex]M_{a}[/tex] + frictional heat gained by [tex]M_{a}[/tex].......(1)
frictional force by [tex]M_{a}[/tex] = [tex]\mu_{k} m_{a} g[/tex]
distance covered = d
KE gained by [tex]M_{b}[/tex] = [tex]o.5m_{b} v^{2}[/tex]
KE gained by [tex]M_{a}[/tex] = [tex]o.5m_{a} v^{2}[/tex]
Frictional heat gained by [tex]M_{a}[/tex] = [tex]\mu_{k} m_{a} gd[/tex]
KE gained by the pulley = [tex]0.5Iw^{2}[/tex] = [tex]0.5 I\frac{v^{2} }{R^{2} }[/tex]
Energy lost by [tex]M_{b}[/tex] due to height = [tex]M_{b}gd[/tex]
Substituting these into equation (1)
[tex]m_{b} gd = 0.5m_{b} v^{2} + 0.5m_{a} v^{2} + \mu_{k} m_{a}gd + 0.6\frac{I}{R^{2} } v^{2}[/tex]
[tex]m_{b} gd - \mu_{k} m_{a}gd = 0.5m_{b} v^{2} + 0.5m_{a} v^{2} + 0.5\frac{I}{R^{2} } v^{2}[/tex]
[tex]m_{b} gd - \mu_{k} m_{a}gd = 0.5m_{b} v^{2} + 0.5m_{a} v^{2} + 0.5\frac{I}{R^{2} } v^{2}\\v^{2} = \frac{m_{b} gd - \mu_{k} m_{a}gd}{0.5m_{b} +0.5m_{a} + \frac{0.5I}{R^{2} } } \\v = \sqrt{\frac{m_{b} gd - \mu_{k} m_{a}gd}{0.5m_{b} +0.5m_{a} + \frac{0.5I}{R^{2} } } }[/tex]
[tex]v = \sqrt{\frac{2(m_{b} gd - \mu_{k} m_{a}gd)}{m_{b} +m_{a} + \frac{I}{R^{2} } } }[/tex]