Derivation of Kinetic Energy Equation

Consider a rope attached to a box of mass m and with initial velocity $v_{i}$. The box is pulled with a constant force F by a rope and undergoes a displacement s, with a final velocity $v_{f}$

Ignoring friction and air resistance and by Conservation of Energy,

$K_{i} + W = K_{f}$, where K is Kinetic energy

$W = K_{f} \, – K_{i}$ $\rightarrow$ Equation 1

Using $v_{f}^{2} = v_{i}^{2} + 2as$

$v_{f}^{2} = v_{i}^{2} + 2 \left( \frac{F}{m} \right) s$, since $F = ma$ by Newton’s 2nd Law

$v_{f}^{2} = v_{i}^{2} + 2\left( \frac{W}{m} \right)$, since W = Fs

$mv_{f}^{2} = m v_{i}^{2} + 2W$

$W = \frac{1}{2}mv_{f}^{2} – \frac{1}{2}m v_{i}^{2}$ $\rightarrow$ Equation 2

Comparing Equation 1 and 2,

$K_{f} = \frac{1}{2} m v_{f}^{2}$ and $K_{i} = \frac{1}{2} m v_{i}^{2}$

We have:

$K = \frac{1}{2} m v^{2}$


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