JEE Main 2025 Apr 3 Shift 1, Physics Q16: Work Energy Theorem And Its Applications

JEE Main2025Apr 3, Shift 1Physics

A particle is released from height $S$ above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.

A
$\frac{S}{2}, \frac{3 g S}{2}$
B
$\frac{S}{2}, \frac{3 g S}{2}$
C
$\frac{S}{4}, \frac{3 g S}{2}$
D
$\frac{S}{4}, \frac{3 g S}{2}$

Solution

Use uniform gravity (near Earth) and energy conservation between release height $S$ and current height $x$ .

$PE at x = m g x, KE at x = \frac{1}{2} m v^{2} .$

Energy conservation: $m g S = m g x + \frac{1}{2} m v^{2} \Rightarrow \frac{1}{2} v^{2} = g (S - x) .$

Condition $KE = 3 PE$ : $\frac{1}{2} m v^{2} = 3 m g x \Rightarrow g (S - x) = 3 g x \Rightarrow x = \frac{S}{4} .$

Speed: $v^{2} = 2 g (S - x) = 2 g (\frac{3 S}{4}) = \frac{3 g S}{2} \Rightarrow v = \frac{3 g S}{2} .$

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JEE Main 2025 Apr 3 Shift 1, Physics Q16: Work Energy Theorem And Its Applications

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