Why Is Mass Halved in the Work-Energy Theorem Derivation?

AI Thread Summary
In the derivation of the work-energy theorem, the mass m is divided by 2 due to the application of the chain rule in calculus. Specifically, the derivative of v² is 2v(dv/dt), which leads to the factor of 1/2 when integrating. This results in the expression for work done being related to the change in kinetic energy. The integration process effectively accounts for the average velocity over the interval, justifying the halving of the mass. Understanding this step is crucial for grasping the relationship between force, work, and energy in physics.
Scalise
Messages
1
Reaction score
0
Hello,

Someone could explain me why in the derivation below the mass m is divided by 2 in the last step?:

##\int\vec{F}\cdot d\vec{s}=m\int\frac{d\vec{v}}{dt}\cdot\vec{v}dt=\frac{m}{2}\int \frac{d}{dt}(v^{2})dt##
 
Physics news on Phys.org
Scalise said:
Hello,

Someone could explain me why in the derivation below the mass m is divided by 2 in the last step?:

##\int\vec{F}\cdot d\vec{s}=m\int\frac{d\vec{v}}{dt}\cdot\vec{v}dt=\frac{m}{2}\int \frac{d}{dt}(v^{2})dt##

The derivitive ##\frac{d}{dt}(v^{2})## is equal to ##2\vec{v}\cdot\frac{d\vec{v}}{dt}##
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Back
Top