Question

A particle released from a height s at certain height kinetic energy is 3 times its potential enerdy

Answer 1

as it is released from height S, its PE = m g S
Let x be the distance through which particle has travelled from the start
Then potential energy = m g (S - x)
By conservation of energy mg S - m g ( S-x) = m g x will be the KE
Given KE = 3 * PE 
So m g x = 3 m g ( S - x)
OR x = 3 S - 3 x
==> 4 x = 3 S
So x = 3/4 * S
Hence the height is S - 3/4 S = 1/4 * S 
Speed = v^2 = 2 g *3/4 S 
So v = (3/2 * g S)^1/2

Answer 2

Given question :-

A particle is released from a height S. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that time are respectively

• $\frac{S}{4},\ \frac{3gs}{4}$
• $\frac{S}{4},\ \frac{\sqrt{3gs} }{4}$
• $\frac{S}{2},\ \frac{\sqrt{3gs} }{2}$
• $\frac{S}{4},\ \sqrt\frac{{3gs} }{2}$

Answer :-

D

Explanation :-

We can realize the situation as shown in fig 1. Let at point C distance x from highest point A, the particle's kinetic energy is three times its potential energy.

Velocity at C,

$\ \ \ \ \ v^2=0+2gh$

$\\or,\ v^2=2gh$ -----------(i)

Potential energy at C

$=mg(S-h)$ ----------(ii)

At point C,

$Kinetic\ energy = 3 \times potential\ energy\\i.e, \ \ \ \frac{1}{2}m\times 2gx=3\times mg(S-x)\\or, \ \ \ \ x=3S-3x\\or, \ \ \ \ 4x=3S\\or, \ \ \ \ S=\frac{4}{3} x\\or, \ \ \ \ x=\frac{3}{4}S$

Therefore, from Eq.(i)

$\ \ \ \ \ v^2=2g\times \frac{3}{4}S\\or,\ v^2=\frac{3}{2}gS\\or,\ v=\sqrt{\frac{3}{2} gS}$

Height of the particle from the ground

$=S-x=S-\frac{3}{4}S=\frac{S}{4}$

Theory :-

Work done by gravity :-

Gravitational potential energy is energy and object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about 9.8 m/s² per second square. Since the zero of gravitational potential energy can be chosen at any point (like the choice of the zero of a co-ordinate system) the potential energy at a height h above that point is equal to the work which would be required to lift the object to that height with no net change in kinetic energy. Since the force required to lift it is equal to its weight it follows that the gravitational potential energy is equal to its weight times height to which it is lifted.

The Work done by gravity formula is given by,

$W_g=-mg(\Delta h)$

where,

m is mass,

g is gravity

h is height

The negative sign indicates that the particle is falling from a height $\Delta h$ vertically in the direction of gravity.

Conservation of mechanical energy :-

According to the principle of conservation of mechanical energy, the mechanical energy of an isolated system remains constant in time, as long as the system is free of friction and other non-conservative forces. In any real situation frictional forces and other non-conservative forces are present, but in many cases their effects on the system are so small that the principle of conservation of mechanical energy can be used as a fair approximation. Though energy cannot be created or destroyed in an isolated system, it can be converted to another form of energy.