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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive x-axis direction is applied to the block . The force $\hat F$ is given by $\hat F=(4-x^2)\hat i \;N$ . Initial position of the block is $x=0$. The maximum kinetic energy of block between $x=0$ to $x=2.0$ is

\[(a)\;5.33\;J \quad (b)\;8.67\;J \quad (c)\; 2.33 \;J \quad (d)\;6.67\;J\]

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1 Answer

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From work energy theorem
Kinetic energy of block
$k=\int \limits_0^x (4-x^2)dx$
$k=4x-\large\frac{x^3}{3}$
For K to be maximum $\large\frac{dh}{dx}$$=0$
$\large\frac{dk}{dx}$$=4-x^2$
$4=x^2$
$x=\pm 2$
At $x=+2 \quad \large\frac{d^2k}{dx^2}$ is -ve . So k is maximum
$k_{max}=4(2)-\large\frac{(2)^3}{3}=\large\frac{16}{3}$$J=5.33\;J$
answered Aug 6, 2013 by meena.p
edited Jun 6, 2014 by lmohan717
 

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