JEE MAIN - Physics (2017 (Offline) - No. 21)
A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is $${1 \over 8}mv_0^2$$, the value of k will be:
10-1 kg m-1 s-1
10-3 kg m-1
10-3 kg s-1
10-4 kg m-1
Explanation
According to the question, final kinetic energy = $${1 \over 8}mv_0^2$$
Let final speed of the body = Vf
So final kinetic energy = $${1 \over 2}mv_f^2$$
According to question,
$${1 \over 2}mv_f^2$$ = $${1 \over 8}mv_0^2$$
$$ \Rightarrow {v_f} = {{{v_0}} \over 2}$$ = $${{10} \over 2}$$ = 5 m/s
Given that, F = –kv2
$$ \Rightarrow $$ $$m\left( {{{dv} \over {dt}}} \right)$$$$ = - k{v^2}$$
$$ \Rightarrow {10^{ - 2}}\left( {{{dv} \over {dt}}} \right) = - k{v^2}$$
$$ \Rightarrow \int\limits_{10}^5 {{{dv} \over {{v^2}}}} = - 100k\int\limits_0^{10} {dt} $$
$$ \Rightarrow {1 \over 5} - {1 \over {10}} = 100k \times 10$$
$$ \Rightarrow k = {10^{ - 4}}kg\,{m^{ - 1}}$$
Let final speed of the body = Vf
So final kinetic energy = $${1 \over 2}mv_f^2$$
According to question,
$${1 \over 2}mv_f^2$$ = $${1 \over 8}mv_0^2$$
$$ \Rightarrow {v_f} = {{{v_0}} \over 2}$$ = $${{10} \over 2}$$ = 5 m/s
Given that, F = –kv2
$$ \Rightarrow $$ $$m\left( {{{dv} \over {dt}}} \right)$$$$ = - k{v^2}$$
$$ \Rightarrow {10^{ - 2}}\left( {{{dv} \over {dt}}} \right) = - k{v^2}$$
$$ \Rightarrow \int\limits_{10}^5 {{{dv} \over {{v^2}}}} = - 100k\int\limits_0^{10} {dt} $$
$$ \Rightarrow {1 \over 5} - {1 \over {10}} = 100k \times 10$$
$$ \Rightarrow k = {10^{ - 4}}kg\,{m^{ - 1}}$$
Comments (0)
