In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1 m s$^{-1}$ at $45^\circ$ from a height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m s$^{-2}$, the kinetic energy of the shotput when it just reaches the ground will be (a) 2.5 J (b) 5.0 J (c) 52.5 J (d) 155.0 J

In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1 m s⁻¹ at 45^° from a

NCERT Exemplar · Q16

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(a) 2.5 J

(b) 5.0 J

(d) 155.0 J

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Concept understanding — Conservation Of Mechanical Energy

Conservation of Mechanical Energy

The Intuition First

Imagine you're holding a heavy stone at shoulder height. Your arm is tired. That stone has potential energy — energy stored because of its position. Now let it go. As it falls, it speeds up. The potential energy is turning into kinetic energy — the energy of motion. Just before it hits the ground, all the original potential energy has become kinetic energy.

Now imagine the reverse: you throw a ball straight up. It leaves your hand fast (lots of kinetic energy), rises, slows down, stops for an instant at the top (zero kinetic energy), then falls back. At the top, all the kinetic energy you gave it has turned back into potential energy.

This back-and-forth transformation — potential ↔ kinetic — is the heart of the idea. Energy doesn't disappear; it just changes form. That's conservation.

The Precise Statement

Important

Conservation of Mechanical Energy: In an isolated system where only conservative forces (like gravity or an ideal spring) do work, the total mechanical energy of the system remains constant.

Total mechanical energy is the sum of kinetic energy ( $K$ ) and potential energy ( $U$ ):

E_{\text{mech}} = K + U

The law says:

K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}}

Or, in symbols:

E_{\text{mech, initial}} = E_{\text{mech, final}}

What This Means in Practice

Let's go back to the falling stone. Suppose you hold it 5 metres above the ground. Its mass is 2 kg. Take $g = 10\ \text{m/s}^2$ .

At the top (initial):

$K_i = 0$ (not moving)

$U_i = mgh = 2 \times 10 \times 5 = 100\ \text{J}$

$E_{\text{mech}} = 0 + 100 = 100\ \text{J}$
Just before hitting ground (final):

$U_f = 0$ (height = 0)

$K_f = \frac{1}{2}mv^2$

Conservation says $K_f = 100\ \text{J}$ , so $\frac{1}{2} \times 2 \times v^2 = 100$ , giving $v = 10\ \text{m/s}$ .

You never needed to know the time of fall or acceleration. Energy conservation gave you the speed directly.

The Two Critical Conditions

Watch out

Mechanical energy is not always conserved. It is conserved only when:

No non-conservative forces (like friction, air resistance, or applied pushes/pulls) do work.
The system is isolated — no external forces transfer energy in or out.

If friction is present, some mechanical energy turns into heat (thermal energy). The total energy of the universe is still conserved, but mechanical energy alone is not.

A Simple Example to Cement It …

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Use conservation of mechanical energy — the shotput’s total energy at launch equals its total energy just before hitting the ground. The kinetic energy on landing is the sum of its initial kinetic energy and the gravitational potential energy it loses. The answer is 155.0 J.

This is a projectile motion problem, but the trick is that you don’t need to calculate the trajectory at all. Air resistance is negligible, so the only force doing work is gravity — a conservative force. That means mechanical energy (kinetic + gravitational potential) is conserved throughout the flight. The shotput starts with some kinetic energy and some gravitational potential energy (because it’s 1.5 m above ground). When it reaches the ground, its potential energy is zero, so all that energy has become kinetic.

Let’s work it through.

Initial kinetic energy Mass $m = 10\ \text{kg}$ , initial speed $u = 1\ \text{m/s}$ .

K_i = \frac{1}{2} m u^2 = \frac{1}{2} \times 10 \times (1)^2 = 5\ \text{J}

Initial gravitational potential energy Height above ground $h = 1.5\ \text{m}$ , $g = 10\ \text{m/s}^2$ .

U_i = m g h = 10 \times 10 \times 1.5 = 150\ \text{J}

Total mechanical energy at launch

E_i = K_i + U_i = 5 + 150 = 155\ \text{J}

Energy just before hitting the ground At ground level, height is zero, so $U_f = 0$ . By conservation of energy,

E_f = E_i = 155\ \text{J}

Since $E_f = K_f + 0$ , we have

$K_f = 155\ \text{J}$ …

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