Question

A body of mass m, moving along the positive x
direction is subjected to a resistive force F = Kv2
(where K is a constant and v the particle velocity).
If m = 10 kg, v = 10 m/s at t = 0, and K = 2 N
(m/s)2 the velocity when t = 2 s is:​

Answer 1

Given:

A body of mass m, moving along the positive x

direction is subjected to a resistive force F = Kv².

(where K is a constant and v the particle velocity).

If m = 10 kg, v = 10 m/s at t = 0, and K = 2 N

(m/s)²

To find:

Velocity at t = 2 sec.

Calculation:

$\therefore \: F = k {v}^{2}$

$= > \: a= \dfrac{ k {v}^{2} }{m}$

$= > \: a= \dfrac{ 2 {v}^{2} }{10}$

$= > \: a= \dfrac{ {v}^{2} }{5}$

$= > \: \dfrac{dv}{dt} = \dfrac{ {v}^{2} }{5}$

$= > \: \dfrac{dv}{ {v}^{2} } = \dfrac{ 1 }{5} \: dt$

Integrating on both sides:

$\displaystyle = > \: \int \dfrac{dv}{ {v}^{2} } = \dfrac{ 1 }{5} \: \int \: dt$

Putting limits:

$\displaystyle = > \: \int_{10}^{v} \dfrac{dv}{ {v}^{2} } = \dfrac{ 1 }{5} \: \int_{0}^{2} \: dt$

$= > \: \bigg \{ - \dfrac{1}{v} \bigg \}_{10}^{v} = \dfrac{ 1 }{5} (2 - 0)$

$= > \: \bigg \{ \dfrac{1}{v} \bigg \}_{v}^{10} = \dfrac{ 2 }{5}$

$= > \: \dfrac{1}{10} - \dfrac{1}{v} = \dfrac{ 2 }{5}$

$= > \: \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{ 2 }{5}$

$= > \: \dfrac{1}{v} = \dfrac{1 - 4}{10}$

$= > \: \dfrac{1}{v} = \dfrac{ - 3}{10}$

$= > \: v = - 3.33 \: m {s}^{ - 1}$

So, final answer is:

$\boxed{ \red{ \bold{ \: v = - 3.33 \: m {s}^{ - 1} }}}$