Work, Energy & Power - Law of Conservation of Mechanical Energy

Vocabulary and key derivation terms for the proof of the law of conservation of mechanical energy in a freely falling body.

Law of conservation of mechanical energy

The law stating that energy can neither be created nor be destroyed, but it can be converted from one form to another form; hence, the total energy remains constant in a closed system.

Potential energy ($P.E$) at Point A

$mgh$, representing the energy at a height $h$ above the ground where the body is initially released.

Kinetic energy ($K.E$) at Point A

$0$, because the initial velocity of the body at release is $v_A = 0$, making $\frac{1}{2}m(0)^2 = 0$.

Total energy ($T.E$) at Point A

$mgh$, calculated by adding potential energy and kinetic energy ($P.E + K.E = mgh + 0$).

Potential energy ($P.E$) at Point B

$mg(h - x) = mgh - mgx$, where $x$ is the distance the body has traveled downward from Point A, leaving it at a height of $h - x$.

Kinetic energy ($K.E$) at Point B

$mgx$, determined by using $v_B^2 - u^2 = 2as$ (with $u = 0$, $a = g$, and $s = x$) to get $v_B^2 = 2gx$, yielding $\frac{1}{2}m(2gx) = mgx$.

Total energy ($T.E$) at Point B

$mgh$, calculated by adding potential energy and kinetic energy: $(mgh - mgx) + mgx = mgh$.

Potential energy ($P.E$) at Point C

$0$, because the body hits the ground where the height is $h = 0$, giving $mg(0) = 0$.

Kinetic energy ($K.E$) at Point C

$mgh$, determined by using $v_C^2 - u^2 = 2as$ (with $u = 0$, $a = g$, and $s = h$) to get $v_C^2 = 2gh$, yielding $\frac{1}{2}m(2gh) = mgh$.

Total energy ($T.E$) at Point C

$mgh$, calculated by adding potential energy and kinetic energy ($P.E + K.E = 0 + mgh$).

Kinematic equation for velocity calculation

$v^2 - u^2 = 2as$, used with $u = 0$ and $a = g$ to find the velocity squared ($v^2$) at different points during free fall.