A body of mass m=10- 2 kg is moving in a medium and experiences a frictional force F = - k v2 . Its initial speed is v0=10ms- 1. After 10s its kinetic energy is (1/8)mv02, then value of k will be:-

Question

A body of mass m=10- 2 kg is moving in a medium and experiences a frictional force F = - k v2 . Its initial speed is v0=10ms- 1. After 10s its kinetic energy is (1/8)mv02, then value of k will be:- (A) 10- 1kgm- 1s- 1 (B) 10- 3kgm- 1 (C) 10- 3kgs- 1 (D) 10- 4kgm- 1

Tardigrade Admin · Accepted Answer

After 10 textsec, (1/2)mv2=(1/8)mv02 ⇒ v=(v0/2)=5ms- 1 Acceleration, a = - (k v2/m) ⇒ (d v/d t) = - (k v2/m) ⇒ displaystyle ∫ 105(d v/v2)=-(k/m) displaystyle ∫ 010dt -(1/5)+(1/10)=(- k/10- 2)× 10 k=10- 4kgm- 1

Q. A body of mass $m = 1 0^{- 2} k g$ is moving in a medium and experiences a frictional force $F = - k v^{2} .$ Its initial speed is $v_{0} = 10 m s^{- 1} .$ After $10 s$ its kinetic energy is $\frac{1}{8} m v_{0}^{2},$ then value of $k$ will be:-

$1 0^{- 1} k g m^{- 1} s^{- 1}$

$1 0^{- 3} k g m^{- 1}$

$1 0^{- 3} k g s^{- 1}$

$1 0^{- 4} k g m^{- 1}$

Q. A body of mass m=10−2kg is moving in a medium and experiences a frictional force F=−kv2. Its initial speed is v0​=10ms−1. After 10s its kinetic energy is 81​mv02​, then value of k will be:-

10−1kgm−1s−1

10−3kgm−1

10−3kgs−1

10−4kgm−1

After 10sec, 21​mv2=81​mv02​ ⇒v=2v0​​=5ms−1 Acceleration, a=−mkv2​ ⇒dtdv​=−mkv2​ ⇒∫105​v2dv​=−mk​∫010​dt −51​+101​=10−2−k​×10 k=10−4kgm−1

Q. A body of mass $m = 1 0^{- 2} k g$ is moving in a medium and experiences a frictional force $F = - k v^{2} .$ Its initial speed is $v_{0} = 10 m s^{- 1} .$ After $10 s$ its kinetic energy is $\frac{1}{8} m v_{0}^{2},$ then value of $k$ will be:-

$1 0^{- 1} k g m^{- 1} s^{- 1}$

$1 0^{- 3} k g m^{- 1}$

$1 0^{- 3} k g s^{- 1}$

$1 0^{- 4} k g m^{- 1}$