Can You Prove Conservation of Energy in Vertical Stone Throwing?

In summary, a stone with weight w is thrown vertically upward into the air from ground level with initial speed v(i). If a constant force f due to air drag acts on the stone throughout its flight, the maximum height reached by the stone is h = v(i)^2 / 2g(1+f/w) and the stone's speed just before impact with the ground is v = v(i) (w-f/w+f)^1/2.
  • #1
jenavira
10
0
This is driving me nuts.

A stone with weight w is thrown vertically upward into the air from ground level with initial speed v(i). If a constant force f due to air drag acts on the stone throughout its flight, (a) show that the maximum height reached by the stone is

h = v(i)^2 / 2g(1+f/w)

and (b) show that the stone's speed just before impact with the ground is

v = v(i) (w-f/w+f)^1/2


I've been working through these for ages and I'm just lost. I know I need the conservation of energy equation 1/2mv^2 = mgh and probably the kinetic motion equation 2a(delta x) = v(f)^2 - v(i)^2, but I'm just not sure what to do with them.
 
Physics news on Phys.org
  • #2
find the stone's acceleration using Newton's 2nd law.
 
  • #3
Huh. I'm getting closer. But I still have no idea how I'm supposed to relate the conservation of energy and kinetic motion equations. (It's probably something really obvious, and I'll feel like an idiot for not realizing it sooner, but I've been staring at these for so long...)
 
  • #4
what is the acceleration that you have got ?
 
  • #5
Well, F = ma so a = F/m...and the force in question would be the initial velocity. But I'm not at all sure how that helps...
 
  • #6
If upward direction is taken to be positve than using Newton's 2nd law you get
-f+(-W)=ma but since W=mg than m=W/g
then you get a = -g(1+f/w) and using this constant acceleration substistute into
[tex]V^2=u^2+2as[/tex] to get the max height. becareful with the sign.
 
  • #7
...right. I'm going to sound like an idiot, but could you go through that accelleration in a bit more detail? I'm still a bit confused.
 
  • #8
[tex]\Sigma\vec{F}=m\vec{a}[/tex] ,right /
two downward forces which are the stone weight and the drag force act on the stone when it is moving upward ,right ?
if the upward direction is chosen to be positive, then downward direction must be negative. then we have -W for the stone weight and -f for the drag force. negative signs indicate the direction and W and f are their magnitudes.
you know that the magnitude of the stone weight is given by W= mg W, m and g are all positive because i am talking about magnitudes.
then substitute into the equation above , you will get what i get.
 
  • #9
I think I've managed it now. Thanks a lot. :)
 

1. How does the law of conservation of energy apply to real-world situations?

The law of conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. This means that in any physical process, the total amount of energy remains constant. In real-world situations, this law is seen in action through various examples such as the conservation of mechanical energy in a swinging pendulum, the conservation of thermal energy in a cup of hot coffee cooling down, or the conservation of chemical energy in a battery powering a device.

2. What is the mathematical proof of the law of conservation of energy?

The mathematical proof of the law of conservation of energy is based on the first law of thermodynamics, which states that the change in internal energy of a closed system is equal to the heat added to the system minus the work done by the system. This can be expressed mathematically as ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added, and W is the work done. By rearranging this equation, we can see that ΔU + W = Q, which means that the total energy in a closed system (ΔU + W) is equal to the heat added (Q), thus proving the conservation of energy.

3. What are some common misconceptions about the law of conservation of energy?

One common misconception about the law of conservation of energy is that it only applies to mechanical energy. In reality, this law applies to all forms of energy, including thermal, chemical, nuclear, and electromagnetic energy. Another misconception is that energy can be created or destroyed. However, as per the law of conservation of energy, energy can only be converted or transferred from one form to another.

4. How does the law of conservation of energy relate to the concept of entropy?

The law of conservation of energy and the concept of entropy are closely related. Entropy is a measure of the amount of disorder or randomness in a system. According to the second law of thermodynamics, the total entropy of a closed system always increases over time, indicating a tendency towards disorder. The law of conservation of energy states that energy can only be transferred or converted, not created or destroyed. Therefore, as energy is converted from one form to another, the overall entropy of the system increases.

5. What are some real-world applications of the law of conservation of energy?

The law of conservation of energy has numerous real-world applications, including in the fields of renewable energy, energy efficiency, and conservation. It is the basis for understanding and harnessing various energy sources such as solar, wind, and hydroelectric power. It also plays a crucial role in designing more efficient machines and processes, reducing energy waste, and finding sustainable solutions for energy consumption. Additionally, this law is essential in understanding the behavior of energy in everyday situations, from cooking food to driving a car.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
770
  • Introductory Physics Homework Help
Replies
15
Views
339
Replies
10
Views
410
  • Introductory Physics Homework Help
Replies
30
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
2
Replies
55
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
Back
Top