Question

Figure shows two blocks in contact sliding down an inclined surface of inclination ${30}^{\circ }.$ The friction coefficient between the block of mass 2.0 kg and the incline is ${\mu }_1=0.20$ and that between the block of mass 4.0 kg and the incline ${\mu }_2=\mathrm{0.30.}$ is Find the acceleration of 2.0 kg block. (g= $10m/s^{-2}$).

• A. $2.7m/s^2$
• B. $2.2m/s^2$
• C. $3.7m/s^2$
• D. $3.2m/s^2$

Answer 1

The correct option is A $2.7m/s^2$

Since, ${\mu }_1<{\mu }_2,$ acceleration of 2 kg-block
down the plane will be more than the acceleration of 4 kg block, if
allowed to move separately. But as the 2.0 kg block is behind the 4.0 kg
block both of them will move with same acceleration say $a$. Taking the
blocks as a system.
Force down the plane on the system $=(4+2)g\sin {30}^{\circ }$
$=\left(6\right)\left(10\right)\left(\frac{1}{2}\right)=30N$
Force up the plane on the system
$={\mu }_1\left(2\right)\left(g\right)\cos {30}^{\circ }+{\mu }_2\left(4\right)\left(g\right)\cos {30}^{\circ }$
$=(2{\mu }_1+4{\mu }_2)g\cos {30}^{\circ }$
$=(2\times 0.2+4\times 0.3)\left(10\right)\left(0.86\right)$
=$13.76$ N
$\therefore$ Net force down the plane is $F=30-13.76=16.24N$
$\therefore$ Acceleration of both the blocks down the plane will be a.
$a=\frac{F}{4+2}=\frac{16.24}{6}=2.7m/s^2$