Proving the Work-Equation Relationship: A Mathematical Analysis

[dan_c_101]

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.

 

[berkeman]

I moved this to the homework forums. Dan, what do you know about kinetic energy KE and how it can be changed?

 

[dan_c_101]

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?

 

[berkeman]

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?

 

[d1v1n1ty nb]

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

 

[Doc Al]

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