Question

Solve the previous problem if the friction coefficient between the 2.0 kg block and the plane below it is 0.5 and the plane below the 4.0 kg block is frictionless.

Answer 1

$m_1=4\text{kg},m_2=2\text{kg}$

$\text{Frictional coefficient between 2 kg block and surface}=0.5$

$R=10\text{cm}=0.1m$

$I=0.5{\text{kg-m}}^2$

$m_{1}gsin\theta -T_1$

$=m_{1}a...\left(1\right)$

$T_2-(m_{2}gsin\theta +\mu m_{2}gcos\theta )+(T_2-T_1)$

$=m_{1}a+m_{2}a$

$=4\times 9.8\times \left(\frac{1}{\sqrt{2}}\right)-$

$\{2\times 9.8\times \left(\frac{1}{\sqrt{2}}\right)+0.5\times 2\times 9.8\times \left(\frac{1}{\sqrt{2}}\right)\}$

$=(4+2+\frac{0.5}{0.01})a$

$\Rightarrow 27.80-(13.90+6.95)=56a$

$\Rightarrow 27.8-20.8=56a$

$\Rightarrow a=\frac{7}{56}=0.125m/s^2$