JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 18)
A body of mass m starts moving from rest along x-axis so that its velocity varies as $$\upsilon = a\sqrt s $$ where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t seconds after the start of the motion is :
$${1 \over 8}\,$$ m a4 t2
8 m a4 t2
4 m a4 t2
$${1 \over 4}\,$$ m a4 t2
Explanation
Given,
$$\upsilon $$ = a $$\sqrt s $$
$$ \Rightarrow $$$$\,\,\,$$ $${{ds} \over {dt}} = a\sqrt s $$
$$ \Rightarrow $$$$\,\,\,$$ $$\int\limits_0^t {{{ds} \over {\sqrt s }}} = \int\limits_0^z {a\,dt} $$
$$ \Rightarrow $$$$\,\,\,$$ 2$$\sqrt s $$ = at
$$ \Rightarrow $$$$\,\,\,$$ s = $${{{a^2}{t^2}} \over 4}$$
= $${1 \over 2}.{{{a^2}} \over 2}.{t^2}$$
$$\therefore\,\,\,\,$$ acceleration = $${{{a^2}} \over 2}$$
$$\therefore\,\,\,$$ Force (F) = m $$ \times $$ $${{{a^2}} \over 2}$$
$$\therefore\,\,\,\,$$ Work done = F. S
= $${{m{a^2}} \over 2} \times {{{a^2}{t^2}} \over 4}$$
= $${{m{a^4}{t^2}} \over 8}$$
$$\upsilon $$ = a $$\sqrt s $$
$$ \Rightarrow $$$$\,\,\,$$ $${{ds} \over {dt}} = a\sqrt s $$
$$ \Rightarrow $$$$\,\,\,$$ $$\int\limits_0^t {{{ds} \over {\sqrt s }}} = \int\limits_0^z {a\,dt} $$
$$ \Rightarrow $$$$\,\,\,$$ 2$$\sqrt s $$ = at
$$ \Rightarrow $$$$\,\,\,$$ s = $${{{a^2}{t^2}} \over 4}$$
= $${1 \over 2}.{{{a^2}} \over 2}.{t^2}$$
$$\therefore\,\,\,\,$$ acceleration = $${{{a^2}} \over 2}$$
$$\therefore\,\,\,$$ Force (F) = m $$ \times $$ $${{{a^2}} \over 2}$$
$$\therefore\,\,\,\,$$ Work done = F. S
= $${{m{a^2}} \over 2} \times {{{a^2}{t^2}} \over 4}$$
= $${{m{a^4}{t^2}} \over 8}$$
Comments (0)
