A body of mass $m = 10^{-2}$ kg is moving in a medium and experiences a frictional force $F = -kv^2$. Its initial speed is $v_0 = 10$ ms$^{-1}$. If, after 10 s, its energy is $\frac{1}{8}mv_0^2$, the value of k will be:

(B) $10^{-1}$ kg m$^{-1}$ s$^{-1}$

JEE Main · 2017 · 02 Apr 2017 · Offline · Work, Power & Energy

A body of mass

m = 1 0^{- 2}

kg is moving in a medium and experiences a frictional force

F = - k v^{2}

. Its initial speed is

v_{0} = 10

^{- 1}

. If, after 10 s, its energy is

\frac{1}{8} m v_{0}^{2}

, the value of k will be:

Answer: (A)

The energy falls to one-eighth, so

\frac{1}{2} m v_{f}^{2} = \frac{1}{8} m v_{0}^{2}

hands us

v_{f} = \frac{v _{0}}{2} = 5

m/s. With the drag

F = - k v^{2}

, Newton's law

m \frac{d v}{d t} = - k v^{2}

separates to

\frac{d v}{v ^{2}} = - \frac{k}{m} d t

, and integrating from

v_{0}

gives

\frac{1}{v} - \frac{1}{v _{0}} = \frac{k}{m} t

. At

t = 10

s this is

\frac{1}{5} - \frac{1}{10} = \frac{k}{1 0 ^{- 2}} (10)

. The right side is

1000 k

(that factor

k / m \cdot t = k /1 0^{- 2} \cdot 10

is the step to handle carefully), so

0.1 = 1000 k

and

k = 1 0^{- 4}

kg m

^{- 1}

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option