Verified Solution • 2025–26 Session moderate

QJEEM18Ph1089: JEE Main 2018 Physics - Class 11 JEEM18Ph1089 Solution: A body of mass m starts moving from rest along x-axis so that…

A body of mass m starts moving from rest along x-axis so that its velocity varies as υ=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 :

Multiple Choice Options

  • A 18 {1 \over 8}\, m a4 t2
  • B 8 m a4 t2
  • C 4 m a4 t2
  • D 14 {1 \over 4}\, m a4 t2

Step-by-Step Verified Solution

Given,

υ\upsilon = a s\sqrt s

⇒\Rightarrow    \,\,\, dsdt=as{{ds} \over {dt}} = a\sqrt s

⇒\Rightarrow    \,\,\, ∫0tdss=∫0za dt\int\limits_0^t {{{ds} \over {\sqrt s }}} = \int\limits_0^z {a\,dt}

⇒\Rightarrow    \,\,\, 2 s\sqrt s = at

⇒\Rightarrow    \,\,\, s = a2t24{{{a^2}{t^2}} \over 4}

= 12.a22.t2{1 \over 2}.{{{a^2}} \over 2}.{t^2}

∴    \therefore\,\,\,\, acceleration = a22{{{a^2}} \over 2}

∴   \therefore\,\,\, Force (F) = m ×\times a22{{{a^2}} \over 2}

∴    \therefore\,\,\,\, Work done = F. S

= ma22×a2t24{{m{a^2}} \over 2} \times {{{a^2}{t^2}} \over 4}

= ma4t28{{m{a^4}{t^2}} \over 8}

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