Question 29: Work, Energy and Power & Circular Motion

Question

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.

Accepted Answer

Using the equation of motion for velocity
at height $ x $:
$    V^2 = 0 + 2g(S - x)    $
Simplifies to:
$    V^2 = 2g(S - x)    $
Potential Energy
at height $ x $:
$    	ext{Potential Energy} = mgx    $
Relation between kinetic and potential energy
:
Given that the kinetic energy is three times the potential energy,
$    mgx = 3 	imes \frac{1}{2} mV^2    $
Simplifying gives:
$    gx = \frac{3}{2} 	imes 2g(S - x)    $
$    gx = 3g(S - x)    $
$    4x = S    $
Solving for $ x $:
$    x = \frac{S}{4}    $
Finding the speed $ V $ at this instant
:
$    V = \sqrt{2g \left( S - \frac{S}{4} ight)} = \sqrt{2g 	imes \frac{3S}{4}} = \sqrt{\frac{3gS}{2}}    $
Thus, the height from the surface of the Earth is $ \frac{S}{4} $ and the speed of the particle at that height is $ \sqrt{\frac{3gS}{2}} $.

Answer

$\frac{\mathrm{S}}{4}, \frac{3 \mathrm{gS}}{2}$

Answer

$\frac{\mathrm{S}}{2}, \frac{3 \mathrm{gS}}{2}$

Answer

$\frac{\mathrm{S}}{4}, \sqrt{\frac{3 \mathrm{gS}}{2}}$

Answer

$\frac{\mathrm{S}}{2}, \sqrt{\frac{3 \mathrm{gS}}{2}}$

Work, Energy and Power & Circular Motion

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.

Choose the correct answer: