A block of weight 100N is pushed by a force F on a horizontal rough plane moves with an acceleration 1ms², when force is doubled its acceleration becomes 10m/s². The coefficient of friction is ______ $\frac{\mathrm{x}}{10}$ . Then the value of ‘$\mathrm{x}$’ is

Now, Let the External Force be F and friction force be f.

Then,

$\begin{array}{l}\mathrm{F}-{\mathrm{f}}_{\mathrm{s}}=\mathrm{ma}\\ \mathrm{F}-\mathrm{\mu N}=\mathrm{ma}\\ \mathrm{F}-\mathrm{\mu }100=10\times 1\\ \mathrm{F}-\mathrm{\mu }100=10\dots .\left(\mathrm{I}\right)\end{array}$

We know, When Force is doubled,

Then,

$\begin{array}{l}2\mathrm{F}-{\mathrm{f}}_{\mathrm{s}}=\mathrm{ma}\\ 2\mathrm{F}-\mathrm{\mu N}=\mathrm{ma}\\ 2\mathrm{F}-\mathrm{\mu }100=10\times 10\\ 2\mathrm{F}-\mathrm{\mu }100=100\dots .\left(\mathrm{II}\right)\end{array}$

Now, from equation (I) and (II)

F=90N

Now, put the value of F in equation (I)

$\begin{array}{l}90-\mathrm{\mu }100=10\\ -\mathrm{\mu }100=10-90\\ \mathrm{\mu }=\frac{80}{100}\\ \mathrm{\mu }=0.8\end{array}$