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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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A body of mass m ascends from the earth's surface with zero initial velocity due to action of two forces as shown. The force $F$ varying with $h$ as $\hat F=-2m \vec {g} (1-ah)$ where a is constant. Find the maximum height atained by the body t

\[(a)\;\frac{1}{ga} \quad (b)\;\frac{1}{a} \quad (c)\;\frac{1}{2a} \quad (d)\;\frac{g}{2a} \]

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Since the body starts from rest and comes to rest at the maximum height the change in kinetic energy of the system is zero and so by work energy theorem total work done is zero
$\int \limits_0^H (\hat F+m \vec g) dh=0$ [Where H is maximum height attained]
$\int \limits_0^H [-2mg(1-ah)+mg]dh=0$
$\int \limits_0^H mg(1-2h)dh=0$
$mg H (1-aH)=0$
$H= \large\frac{1}{a}$
Maximum height attained $=\large\frac{1}{a}$
Hence b is the correct answer.
answered Aug 29, 2013 by meena.p
edited Jul 3, 2014 by lmohan717

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