Work Power And Energy Ques 35
- A body of mass $m=10^{-2} kg$ is moving in a medium and experiences a frictional force $F=-k v^{2}$. Its initial speed is $v _0=10 ms^{-1}$. If, after $10 s$, its energy is $\frac{1}{8} m v _0^{2}$, the value of $k$ will be
(2017 Main)
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Answer:
Correct Answer: 35.(b)
Solution:
Formula:
- Given, force, $F=-k v^{2}$
$\therefore$ Acceleration, $a=\frac{-k}{m} v^{2}$
$$ \begin{aligned} \text { or } & & \frac{d v}{d t} & =\frac{-k}{m} v^{2} \\ \Rightarrow & & \frac{d v}{v^{2}} & =-\frac{k}{m} \cdot d t \end{aligned} $$
Now, with limits, we have
$$ \begin{aligned} & \int _{10}^{v} \frac{d v}{v^{2}}=-\frac{k}{m} \int _0^{t} d t \\ & \Rightarrow \quad-\frac{1}{v} _{10}^{v}=-\frac{k}{m} t \\ & \Rightarrow \quad \frac{1}{v}=0.1+\frac{k t}{m} \\ & \Rightarrow \quad v=\frac{1}{0.1+\frac{k t}{m}}=\frac{1}{0.1+1000 k} \\ & \Rightarrow \quad \frac{1}{2} \times m \times v^{2}=\frac{1}{8} \times v _0^{2} \\ & \Rightarrow \quad v=\frac{v _0}{2}=5 \\ & \Rightarrow \quad \frac{1}{0.1+1000 k}=5 \\ & \Rightarrow \quad 1=0.5+5000 k \\ & \Rightarrow \quad k=\frac{0.5}{5000} \quad \Rightarrow \quad k=10^{-4} kg / m \end{aligned} $$
BETA
