Question

A body of mass $m=1$ kg is moving in a medium and experiences a frictional force $F=-kv$, where $v$ is the speed of the body. The initial speed is $v_0=10\,\text{ms}^{-1}$ and after $10$ s, its energy becomes half of initial energy. Then, the value of $k$ is

• A. $10 \ln \sqrt{2}$
• B. $\ln \sqrt{2}$
• C. $\frac{\ln 2}{20}$
• D. $10 \ln 2$
• E. $\ln 2$

Hint:

If energy becomes half, velocity becomes $\frac{1}{\sqrt{2}}$ of its initial value. This is a common shortcut in exponential decay problems.

Answer 1

The Correct Option is C

Concept:
Force $F = ma = m \frac{dv}{dt}$. For $F = -kv$, we have a differential equation for velocity. Kinetic Energy $E = \frac{1}{2}mv^2$.

Step 1: Solve for Velocity $v(t)$.
\[ m \frac{dv}{dt} = -kv \implies \frac{dv}{v} = -\frac{k}{m}dt \] Integrating from $v_0$ to $v$ and $0$ to $t$: \[ \ln\left(\frac{v}{v_0}\right) = -\frac{kt}{m} \implies v = v_0 e^{-kt/m} \]

Step 2: Use the Energy Condition.
Energy $E \propto v^2$. Given $E_{final} = \frac{1}{2}E_{initial}$: \[ \frac{1}{2}m v^2 = \frac{1}{2} \left( \frac{1}{2}m v_0^2 \right) \implies v^2 = \frac{v_0^2}{2} \implies v = \frac{v_0}{\sqrt{2}} \]

Step 3: Substitute values to find $k$.
At $t = 10$ s and $m = 1$ kg: \[ \frac{v_0}{\sqrt{2}} = v_0 e^{-k(10)/1} \implies \frac{1}{\sqrt{2}} = e^{-10k} \] Taking natural log: \[ \ln(2^{-1/2}) = -10k \implies -\frac{1}{2}\ln 2 = -10k \] \[ k = \frac{\ln 2}{20} \]