Question

A particle moves on a rough horizontal ground with some initial velocity say v₀. If (3/4)th of its kinetic energy is lost in friction in time t₀, then coefficient of friction between the particle and the ground is.

• A. $v_0/2g{t}_0$
• B. $v_0/4g{t}_0$
• C. $3v_0/4g{t}_0$
• D. $v_0/g{t}_0$

Answer 1

The correct option is A

$v_0/2g{t}_0$

Explanation for the correct option

Step 1. Given data

Initial velocity $=v_0$

Time $=t_0$

Kinetic energy lost $=3/4oftotalkineticenergy$

Step 2. Assumptions

Mass $=m_0$

Coefficient of friction $=\mu$

Acceleration $=a$

Final velocity $=v$

Step 3. Calculation of coefficient of friction

Initial kinetic energy $=1/2m{v^2}_0$

Final kinetic energy $=1/2m{v}^2$……..(a)

Final kinetic energy $1/2m{v}^{2^{}}=(1/2m{v^2}_0)-(3/4\times 1/2m{v^2}_0)$………….(b)

From equations (a) and (b), we get.

$1/2m{v}^2=(1/2m{v^2}_0)-(3/4\times 1/2m{v^2}_0)$

$$\begin{gathered} 1/2m{v}^2=(4m{v^2}_0-3m{v_0}^2)/8 \\ 1/2m{v}^2=m{v^2}_0/8 \\ {v}^2={v^2}_0/4 \\ v=v_0/2 \end{gathered}$$

Using, $v=v_0+at$

$$\begin{gathered} v_0/2=v_0+at \\ at=v_0-v_0/2 \\ a=v_0/2t_0 \end{gathered}$$

Using, $a=\mu g$

$\mu =a/g$

Using,

$a=v_0/2t_0$,

we get.

$\mu =v_0/2g{t}_0$

Hence, option (a) will be correct.