Question

Two blocks connected by a massless string slide down an inclined plane having angle of inclination as ${37}^{\circ }$. The masses of two blocks are $4$ $kg$ and $2$ $kg$ with $\mu$ as $0.75$ and $0.25$ respectively. Then______(Take $g=9.8m/s^2$)

• A. The common acceleration of two masses is $1.3{ms}^{-2}$ and tension in the string is $5.4N$.
• B. The tension in the string is $14.9N$.
• C. The acceleration of the masses is $1.5$ $m{s}^{-2}$
  
• D. The acceleration of masses is $5.3$ ${ms}^{-2}$ and tension in the string is $1.3$ $N$.

Answer 1

The correct option is A The common acceleration of two masses is $1.3{ms}^{-2}$ and tension in the string is $5.4N$.

From FBD of 2 kg block,
$2a=2g\sin 37-T-f$
$2a=2g\sin 37-T-0.25\left(2g\cos 37\right)............................\left(1\right)$
From FBD of 4 kg block,
$4a=4g\sin 37+T-f^{\prime }$
$4a=4g\sin 37+T-0.75\left(4g\cos 37\right)............................\left(2\right)$
Eq.$\left(1\right)$+Eq. $\left(2\right),$
$6a=6g\sin 37-3.5g\cos 37$
$\Rightarrow a=1.3m/s^2$
put the value in (1), we will get,
$2.6=12-T-4$
$\Rightarrow T=8-2.6=5.4N$