Conservation of energy for toy gun

AI Thread Summary
In a spring-loaded toy gun scenario, mechanical energy is conserved because no nonconservative forces, such as air resistance, act on the ball after it is released. The gravitational potential energy of the ball is calculated using mgh, while the spring's potential energy is given by 1/2 k x^2. The maximum height the ball reaches can be derived from the energy stored in the spring. The spring's potential energy is proportional to the square of its compression distance and inversely related to gravitational effects. The discussion highlights the importance of understanding potential energy in relation to both the spring and gravitational forces.
WY
Messages
28
Reaction score
0
I've been reading about a situation on the conservation of energy:
A spring-loaded toy gun is used to shoot a ball of mass m straight up in the air. View Figure The spring has spring constant k. If the spring is compressed a distance x_0 from its equilibrium position and then released, the ball reaches a maximum height h_max (measured from the equilibrium position of the spring). There is no air resistance, and the ball never touches the inside of the gun. Assume x_o is greater than h

Now my question to this is that:
Is mechanical energy conserved because no nonconservative forces perform work on the ball and do nonconservative forces act in this situation after the ball is released at all? and do the forces of gravity and the spring have potential energies associated with them?

As you can tell I don't really have a great grasp on these concepts, so would anyone like to enlighten me please? thanks so much!
 
Physics news on Phys.org
WY said:
Now my question to this is that:
Is mechanical energy conserved because no nonconservative forces perform work on the ball and do nonconservative forces act in this situation after the ball is released at all?
No nonconservative forces act on the ball at any time (in this problem). That's why they specify "There is no air resistance, and the ball never touches the inside of the gun."; those would be nonconservative forces.

and do the forces of gravity and the spring have potential energies associated with them?
Absolutely. Gravitational potential energy (near the Earth's surface) is given by mgh, where "h" is height measured from some arbitrary reference point. Spring potential energy is given by 1/2 k x^2, where x is the displacement from the unstretched position.
 
so, h,max=(1/2)kx^2/gm
look closely at this equation and it will make sense to you!
the stiffer the spring the larger the "k" the more energy the spring has when it is compressed, and the higher the ball will go for a given x the distance the spring is compressed! A springs potential energy stored is proportional [1/2 k] to its distance compressed squared.
and inversely proportional to both the Earth's gravitational constant and the mass of the object shot up!

this should make some sense, right?

love and peace,
and,
peace and love,
(kirk) kirk gregory czuhai
http://www.altelco.net/~lovekgc/kirksresume.htm
 
Last edited by a moderator:
Kirk,

"A springs potential energy stored is proportional [1/2 k] to its distance compressed squared. and inversely proportional to both the Earth's gravitational constant and the mass of the object shot up! this should make some sense, right?"

Actually no.

What does "the Earth's gravitational constant" or "the mass of the object shot up" have to do with a spring's potential energy?
 
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...
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?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top