Proving the Work-Equation Relationship: A Mathematical Analysis

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The discussion centers on proving the relationship between work done (W) and the change in kinetic energy (Ek) when a mass (m) accelerates uniformly from an initial velocity (v) to a final velocity (v'). The equation W = Ek = mv'^2/2 - mv^2/2 is highlighted, emphasizing the need to understand how kinetic energy is derived from mass and velocity. Participants express confusion about how to mathematically demonstrate this relationship, particularly in terms of applied force and energy changes. The conversation also touches on momentum and energy loss in collision scenarios, illustrating the principles with examples. Ultimately, the thread aims to clarify the integration of force and displacement to derive the work-energy theorem.
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Prove mathematically that if a mass (m) speeds up uniformly from v to v’ while being acted upon by an applied force (Fap) in the same direction as its displacement , then W = Ek = mv'^2/2 - mv^2/2.
 
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I moved this to the homework forums. Dan, what do you know about kinetic energy KE and how it can be changed?
 
You get kinetic energy from mass x velocity^2 / 2. I am not sure how it can be changed besides the mass and velocity. Can you explain? Sorry the equation parts just confuse me, but can you help me somewhat to show how mv'^2/2 - mv^2/2 can equal the work done?
 
It is true that KE = \frac{m v^2}{2}

But how did the object get its velocity? Who does the work on a baseball to get it moving with its velocity? What was the KE of the baseball before being thrown?
 
This sounds to me like your being introduced into momentum in terms of energy.you would need the Mass and Velocity of the applied force.

Example: Car Crashes into the back of another car (Fap being mass 1)
Car behind: Mass 1 (m1) Car in front: Mass 2 (m2)
Before Collision After Collision
Initial Velocity of Mass 1 (u1) Final Velocity of Mass 1 (V1)
Initial Velocity of Mass 2 (u2) Final Velocity os Mass 2 (V2)

So: 1/2(m1u1^2)+1/2(m2u^2)=1/2(m1V1^2)+1/2(m2V2^2)+Energy lost
Therefore: 1/2(m1u1^2+m2u2^2)-1/2(m1V1^2+m2V2^2)= Energy Lost
 
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dan_c_101 said:
Prove mathematically that if a mass (m) speeds up uniformly from v to v’ while being acted upon by an applied force (Fap) in the same direction as its displacement , then W = Ek = mv'^2/2 - mv^2/2.
I suspect you are being asked to show that:
\int F \cdot ds = \Delta{KE}

Hint: Make use of the fact that:
F = m \frac{dv}{dt}

to change the variable of integration
 
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