In the figure shown, the frictional coefficient between table and block is 0.2. The ratio of tensions in the right and left strings is $\frac{17}{p+17}$ .The value of $p$ is_______ (Use $g=10m{s}^{-2}$ )

 

Net pulling force $F_1=15g-5g$

$F_1=10g=100N$

Net stopping force $F_2=0.2\times 5\times 10=10N$

Since $F_1>F_2,$ therefore the system will move (anti-clockwise) with an acceleration,  $a=\frac{F_1-F_2}{15+5+5}=\frac{90}{25}= 3.6 m/s^2$

$$\begin{gathered} T_1-5g=5\left(3.6\right) \\ \Rightarrow T_1=68N \end{gathered}$$

$$\begin{gathered} 15g-T_2=15\left(3.6\right) \\ \Rightarrow T_2=96N \end{gathered}$$

Ratio of tensions, $\frac{T_1}{T_2}=\frac{68}{96}=\frac{17}{24}=\frac{17}{7+17}$