Question

A block moving on a surface with a velocity of $20m{s}^{-1}$ comes to rest because of surface friction over a distance of $40m$. The coefficient of dynamic friction is (take, $g=10m{s}^{-2}$)

• A. $0.5$
• B. $0.3$
• C. $0.2$
• D. $0.1$

Answer 1

The correct option is A

$0.5$

Step 1: Defining the coefficient of friction

1. It is the ratio between the frictional force opposing the motion of two surfaces which are in contact, to the normal force existing between the two surfaces.
2. It is a unit less quantity and is denoted by Greek letter $\text{'}\mu \text{'}$. Mathematically it is equal to:

$$\begin{gathered} \mu =\frac{F}{N}=\frac{ma}{mg}=\frac{a}{g} \\ m=massofbody \\ a=accelerationofbody \\ g=accelerationduetogravity=10m{s}^{-2} \end{gathered}$$

Step 2: Given data

The initial velocity of a body, $u=20m{s}^{-1}$

The final velocity of a body, $v=0\because bodycomestorestduetofriction$

The distance covered by a body, $S=40m$

Step 3: Calculating the coefficient of dynamic friction

Using the third equation of motion, we get

$$\begin{gathered} v^2=u^2+2aS \\ \Rightarrow 0={\left(20\right)}^2-2\times \left(\mu g\right)S\because (a=\mu g)isinoppositedirection \\ \Rightarrow 0=400-2\times 10\times 40\times \mu \\ \Rightarrow \mu =\frac{400}{800} \\ \Rightarrow \mu =0.5 \end{gathered}$$

Hence, option(A) is correct answer.