Physics: State the work-energy theorem

E=h.f =5.2×1014×6.63×10^-34E=3.496×10^-30 J...

State the work-energy theorem - id15254507, LoLGuy133

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mmunozgo9334

05.02.2022

Physics

Work Energy Theorem :-It states that net work done on any body is equal to the change in its kinetic energy .

We could derive this , as ;

Consider a body of mass m being pushed by a force F acting along the horizontal , due to which it is displaced s m away . Since the angle between the force and the displacement is 0° , work done will be ,

$\sf \longrightarrow Work = F s cos\theta \\$

$\sf \longrightarrow Work = (ma)(s)(cos0^o)\\$

$\sf \longrightarrow\pink{ Work = m \ a \ s } \dots (i)$

Now let's use the third equation of motion namely,

$\sf \longrightarrow 2as = v^2 -u^2$

where the symbols have their usual meaning.

$\sf \longrightarrow as =\dfrac{1}{2}(v - u)^2\\$

Multiplying both sides by m,

$\sf \longrightarrow mas = \dfrac{m}{2}(v-u)^2$

Now from equation (i),

$\sf \longrightarrow Work = \underbrace{\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2}$

Above term on RHS is change in the Kinetic energy , therefore ,

$\sf \longrightarrow \underline{\boxed{\bf Work = \Delta Energy_{(Kinetic)} }}$

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State the work-energy theorem

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