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Question: A block of weight \(100N\) is pushed by a force \(F\) on a horizontal rough plane moves with an acce...

A block of weight 100N100N is pushed by a force FF on a horizontal rough plane moves with an acceleration 1m/s21m/{s^2}, when force is doubled its acceleration becomes 10m/s210m/{s^2}. The coefficient of friction is____(g=10m/s2)\left( {g = 10m/{s^2}} \right)
\eqalign{ & \left( {\text{A}} \right){\text{ }}0.2 \cr & \left( {\text{B}} \right){\text{ }}0.4 \cr & \left( {\text{C}} \right){\text{ }}0.8 \cr & \left( {\text{D}} \right){\text{ }}0.5 \cr}

Explanation

Solution

Firstly calculate the mathematical expressions for the force when then the only force is applied on the block, after that calculate the force equation when the force becomes double. Then solve both equations and then find the value of the coefficient.
Formula used:
F−fs=maF - {f_s} = ma

Complete answer:
The given setup can be completely understood by the following diagram:

Given that,
Weight, W=100NW = 100N
Acceleration, a=1m/s2a = 1m/{s^2}
Acceleration when force is double, a=10m/s2a = 10m/{s^2}
Now, Let the External Force be FF and frictional force be ff
Then,
\matrix  F−fs=ma   ⇒F−μN=ma [∵fs=μN]   ⇒F−μ100=10×1   ⇒F−μ100=10....(I) \matrix {\;F - {f_s} = ma} \\\ {\; \Rightarrow F - \mu N = ma{\text{ }}\left[ {\because {f_s} = \mu N} \right]} \\\ {\; \Rightarrow F - \mu 100 = 10 \times 1} \\\ {\; \Rightarrow F - \mu 100 = 10....\left( I \right)} \\\
We know, When Force is doubled,
Then,
\matrix2F−fs=ma ⇒2F−μN=ma ⇒2F−μ100=10×10 ⇒2F−μ100=100....(II)    \matrix {2F - {f_s} = ma} \\\ { \Rightarrow 2F - \mu N = ma} \\\ { \Rightarrow 2F - \mu 100 = 10 \times 10} \\\ { \Rightarrow 2F - \mu 100 = 100....\left( {II} \right)} \\\ \; \\\
Now, from equation (I)\left( I \right) and (II)\left( {II} \right)
F=90NF = 90N
Now, put the value of FF in equation (I)\left( I \right)
\matrix90−μ100=10 ⇒−μ100=10−90 ⇒μ=80100 ∴μ=0.8 \matrix {90 - \mu 100 = 10} \\\ { \Rightarrow - \mu 100 = 10 - 90} \\\ { \Rightarrow \mu = \dfrac{{80}}{{100}}} \\\ {\therefore \mu = 0.8} \\\
Hence, the coefficient of restitution is 0.80.8.

So, the correct option is C.

Additional Information:
The coefficient of friction is defined because of the ratio of the force required to maneuver two sliding surfaces over one another, and therefore the force holding them together. When surfaces in touch move relative to every other, the friction between the 2 surfaces converts K.E. into thermal energy (that is, it converts work to heat). This property can have dramatic consequences, as illustrated by the utilization of friction created by rubbing pieces of wood together to start out a fireplace. K.E. is converted to thermal energy whenever motion with friction occurs.

Note:
Friction isn't itself a fundamental force. Dry friction arises from a mixture of inter-surface adhesion, surface roughness, surface deformation, and surface contamination. Friction may be a non-conservative force - work done against friction is path-dependent. Within the presence of friction, some K.E. is usually transformed into thermal energy, so energy isn't conserved.