Question

prove that the total work done on a particle is equal to the change in it's kinetic energy​

Answer 1

Answer:

$\underline{\bold{ \mathfrak{WORK-ENERGY \ THEOREM}}}$

As we know, Kinetic Energy (K):

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

Rate of change of Kinetic Energy with time is:

$\sf \frac{dK}{dt} = \frac{d}{dt} ( \frac{1}{2} m {v}^{2} ) \\ \\ \sf = \frac{1}{2} m \frac{d}{dt} ( {v}^{2} ) \: \: \: \: \: \: ...(as \: m \: is \: constant) \\ \\ \sf = \frac{1}{2} m \frac{d( {v}^{2}) }{dt} \times \frac{dv}{dv} \\ \\ \sf = \frac{1}{2} m \frac{d ({v}^{2} )}{dv} \times \frac{dv}{dt} \\ \\ \sf = \frac{1}{ \cancel{2}} m \times \cancel{2}v \frac{dv}{dt} \\ \\ \sf = mv \frac{dv}{dt} \\ \\ \sf = mav \: \: \: \: \: \: ...( \frac{dv}{dt} = a) \\ \\ \sf = Fv \: \: \: \: \: ...(F = ma) \\ \\ \sf = F \frac{dx}{dt} \: \: \: \: \: \: ...(v = \frac{dx}{dt} )$

$\sf \implies \frac{dK}{ \cancel{dt}} = F \frac{dx}{ \cancel{dt}} \\ \\ \sf \implies \int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} Fdx \\ \\ \sf \implies W = K_f - K_i \: \: \: \: \: \: ...(\int_{x_i}^{x_f} Fdx = W)$

(Where $\sf K_f \ and K_i$ are final and initial kinetic energy corresponding to final $\sf x_f$ and initial $\sf x_i$ position respectively.)

$\boxed{ \bold{ \huge{W = \Delta K}}}$

Answer 2

Answer:

v² = u² + 2as

v² - u² = 2as

mv² - mu² = 2mas

1/2mv² - 1/2mu² = Ga

Final Kinetic energy - Initial Kinetic energy = Work done

W = ΔK