Name: Potential Energy of an Object at a Height
Uploaded: 2019-07-19
Duration: 8 min 3 s
Description: Gravitational potential energy is the energy stored in an object due to its position. It depends upon the height at which the object is placed. Let us derive...

Question

How can you show that the sum "

K . E + P . E

" is the same for all positions in the freely falling of a body?

Accepted Answer

Suppose a body of mass m is raised at a vertical height $h$ above the ground at A. The body is kept at rest. So, it has only potential energy. As the body falls vertically downwards, its height decreases so its potential energy decreases. At the same time, its velocity increases so its kinetic energy increases. According to the law of conservation of mechanical energy, the decrease in potential energy is exactly equal to the increase in kinetic energy or the sum of these two energies at any point is a constant.

1. At position A, $P . E + K . E = m g h, K . E = 0$

Therefore, $T o t a l E n e g y = P . E + K . E = m g h + 0 = m g h$

2. At position B, It is at a depth $x$ below $A$ or at a height $(h - x)$ above the ground.

Therefore, $P . E = m g (h - x)$

Suppose $v_{1}$ be the velocity at $B$ .

For the falling from $A$ to $B$ , we get from the kinematic equation:

$v^{2} = u^{2} + 2 a s \to {v_{1}}^{2} = 0 + 2 (+ g) . x [A s u = 0, v = v_{1}, a = + g]$

or, ${v_{1}}^{2} = 2 g x$

Therefore, $K . E = \frac{1}{2} m {v_{1}}^{2} = \frac{1}{2} . m . 2 g x = m g x$

Therefore, $T o t a l E n e r g y = m g (h - x) + m g x = m g h - m g x + m g x = m g h$

3. Position C is a point, just before touching the ground.

Here, vertical height, $h = 0$ , Therefore $P . E = 0$

If $V_{2}$ be the velocity at $C$ then from $v^{2} = u^{2} + 2 a s$ we get,

${v_{2}}^{2} = 0 + 2 g h = 2 g h$

Therefore, $K . E = \frac{1}{2} m {v_{2}}^{2} = \frac{1}{2} . 2 g h = m g h$

Therefore, $T o t a l E n e r g y = 0 + m g h = m g h$

Thus, the total energy remains constant at all points along the path of the body.

How can you show that the sum " $K . E + P . E$ " is the same for all positions in the freely falling of a body?

Important Questions on Effect of Energy : Work, Power & Energy

Learn and Prepare for any exam you want !

How can you show that the sum "K.E+P.E" is the same for all positions in the freely falling of a body?

Important Questions on Effect of Energy : Work, Power & Energy

Learn and Prepare for any exam you want !

How can you show that the sum " $K . E + P . E$ " is the same for all positions in the freely falling of a body?