- #1
n3pix
- 15
- 1
Hello,
I'm confusing about the basic terms about Conservation of Energy, Potential Energy and Work.
Consider that we have a mass ##M## above the ground (zero point) distance of ##y_{0}=h##. When we release the mass it will accelerate through it's way to ground. So the work is made by a field force (gravitation) that is ##F_{G}=-Mg##. The last position of the mass is ##y=0##. So the displacement of mass is equal to ##(y-y_{0})=(0-h)=-h##. Therefore, the work done by gravity is ##W(by gravity)=F_{G}\times(y-y_{0})=(-Mg)(-h)=Mgh##. As the definition of work (as I learnt), the work done by something is equal to the force times displacement of the body and It's equal to change in kinetic energy of that body. So, if we consider that the mass has initial velocity ##V_{0}=0## and last velocity at the moment it touch the ground ##V##. Therefore, work done by gravity is equal to ##W(by gravity)=Mgh=\Delta{KE}=\frac{1}{2}M{{V}^{2}}-\frac{1}{2}M{{V_{0}}^{2}}=\frac{1}{2}M{V^{2}}-\frac{1}{2}M\times{0}=\frac{1}{2}M{V^{2}}##.
I hope there is no problem about my ubderstanding about this example (falling body example).
I will now give you reverse example, mass M raised by us. To raise mass M from the ground or heigh of ##y_{0}=0## to height of ##y=h## without acceleration we must apply an upward, equal and opposite force ##F_{ag}=-F_{G}=Mg##. So the work done by us is ##W(by us)=F_{ag}(y-y_{0})=(Mg)(h-0)=Mgh##. As the definition of work, there must be change in kinetic energy of the object. The initial velocity and last velocity of the object are same and the mass has no acceleration. Therefore, ##W(by us)=Mgh=\Delta{KE}=\frac{1}{2}M{{V}^{2}}-\frac{1}{2}M{{V}^{2}}=0##.
What does this mean?
I'm confusing about the basic terms about Conservation of Energy, Potential Energy and Work.
Consider that we have a mass ##M## above the ground (zero point) distance of ##y_{0}=h##. When we release the mass it will accelerate through it's way to ground. So the work is made by a field force (gravitation) that is ##F_{G}=-Mg##. The last position of the mass is ##y=0##. So the displacement of mass is equal to ##(y-y_{0})=(0-h)=-h##. Therefore, the work done by gravity is ##W(by gravity)=F_{G}\times(y-y_{0})=(-Mg)(-h)=Mgh##. As the definition of work (as I learnt), the work done by something is equal to the force times displacement of the body and It's equal to change in kinetic energy of that body. So, if we consider that the mass has initial velocity ##V_{0}=0## and last velocity at the moment it touch the ground ##V##. Therefore, work done by gravity is equal to ##W(by gravity)=Mgh=\Delta{KE}=\frac{1}{2}M{{V}^{2}}-\frac{1}{2}M{{V_{0}}^{2}}=\frac{1}{2}M{V^{2}}-\frac{1}{2}M\times{0}=\frac{1}{2}M{V^{2}}##.
I hope there is no problem about my ubderstanding about this example (falling body example).
I will now give you reverse example, mass M raised by us. To raise mass M from the ground or heigh of ##y_{0}=0## to height of ##y=h## without acceleration we must apply an upward, equal and opposite force ##F_{ag}=-F_{G}=Mg##. So the work done by us is ##W(by us)=F_{ag}(y-y_{0})=(Mg)(h-0)=Mgh##. As the definition of work, there must be change in kinetic energy of the object. The initial velocity and last velocity of the object are same and the mass has no acceleration. Therefore, ##W(by us)=Mgh=\Delta{KE}=\frac{1}{2}M{{V}^{2}}-\frac{1}{2}M{{V}^{2}}=0##.
What does this mean?
