Question

Figure shows two blocks in contact sliding down an inclined surface of inclination ${30}^0$. The friction coefficient between the block of mass 2 kg and the incline is ${\mu }_1$ = 0.20 and that between the block of mass 4 kg and the incline is ${\mu }_2$=0.30. Find the acceleration of 2.0 kg block. $g=10m{s}^{-2}$.

Answer 1

Since, $\mu <{\mu }_2$, acceleration of 2-kg block down the plane will be more the acceleration of 4-kg block, if allowed to move separately.

In this case, both blocks are treated as a system of mass (4+2) = 6kg and will move down with the same acceleration.

Net force down the plane is
$F=(m_1+m_2)gsin\theta -{\mu }_1{m}_{1}gcos\theta -{\mu }_2{m}_{2}gcos\theta$

$=(4+2)gsin{30}^0-\left(0.2\right)\left(2\right)gcos{30}^0-\left(0.3\right)\left(4\right)gcos{30}^0$

$=\left(6\right)\left(10\right)\left(\frac{1}{2}\right)-\left(0.4\right)\left(10\right)\left(\frac{\sqrt{3}}{2}\right)-\left(1.2\right)\left(10\right)\left(\frac{\sqrt{3}}{2}\right)$

$=30-13.76=16.24N$

Therefore, acceleration of both the blocks down the plane will be
$a=\frac{F}{m_1+m_2}$

$=\frac{16.24}{4+2}=2.7m{s}^{-2}$