Question

In the figure shown the friction between the $4$ kg block and the incline as ${\mu }_1$ and between $8$ kg and incline is ${\mu }_2$. (Take $g=10m/s^2$) If ${\mu }_1=0.2$ and ${\mu }_2=0.3$ then find acceleration of $m_1$ and $m_2$?

• A. $a_1=a_2=2.4m/s^2$
• B. $a_1=3.2m/s^2,a_2=2.4m/s^2$
• C. $a_1=a_2=3.2m/s^2$
• D. $a_1=2.4m/s^2,a_2=3.2m/s^2$

Answer 1

The correct option is A $a_1=a_2=2.4m/s^2$

$N=mg\cos {30}^o$

$F_{net}=mg\sin \theta =\mu mg\cos \theta$

$a=\frac{F}{m}=g\sin {30}^o-\mu g\cos {30}^o$

$=10-\frac{1}{2}-\frac{2}{10}\times \sqrt{10}\times \frac{\sqrt{3}}{2}$

$=5-\sqrt{3}=3.267$

$a=g\sin {30}^o-{\mu }_{2}g\cos {30}^o$

$=5-\frac{3}{10}\times 10\times \frac{\sqrt{3}}{2}$

$=5-\frac{3\sqrt{3}}{2}=2.40$

As acceleration of $8kg$ block is less than that of $4kg$ block. Both the blocks will have same acceleration which is equal to $2.4m/s^2$

$a_1=a_2=2.4m/s^2$