A body of mass m = 10- 2 textkg is moving in a medium and experiences a frictional force F = - k v2 . Its initial speed is v0 = 10 textms- 1 . After 10 texts, its kinetic energy is (1/8) textmv02 , then value of k will be:-

Question

A body of mass m = 10- 2 textkg is moving in a medium and experiences a frictional force F = - k v2 . Its initial speed is v0 = 10 textms- 1 . After 10 texts, its kinetic energy is (1/8) textmv02 , then value of k will be:- (A) 10- 1k textg textm- 1 texts- 1 (B) 10- 3 text kg textm- 1 (C) 10- 3 textkg s- 1 (D) 10- 4 textkg textm- 1

Tardigrade Admin · Accepted Answer

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

Q. 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 $v_{0} = 10 ms^{- 1} .$ After $10 s,$ its kinetic energy is $\frac{1}{8} mv_{0}^{2},$ then value of $k$ will be:-

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

$1 0^{- 3} kg m^{- 1}$

$1 0^{- 3} kg s^{- 1}$

$1 0^{- 4} kg m^{- 1}$

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:-

10−1kg m−1s−1

10−3 kg m−1

10−3kg s−1

10−4kgm−1

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

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

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

$1 0^{- 3} kg m^{- 1}$

$1 0^{- 3} kg s^{- 1}$

$1 0^{- 4} kg m^{- 1}$