- #1
failedphysics
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I really don't understand this example:
A feather or spring or whatever it's called got the stiffness 29.5 N/m. It is lying inside a horizontal tube (feather cannon). The feather /spring is connected in the very end of the tube. The feather / pring is as "long" as the tube if it's not being pressured longer into the tube or stretched out. We press the feather / spring 12.5cm in and then locks it there. Then we unlocks it and a ball gets shoot out.
(From now on I will call the spring or feather or whatever it is a feather, not sure if it's correctly used)
a) How big is the speed of the feather when it passes the end of the tube?
b) What will be the speed of the ball if we put the tube vertical and shoot the ball up?
Solution: When the feather is tense, it has potential energy: Ep = 1/2kx^2 = 1/2 * 29.5 N /m * (0.125m)^2 = 0.230 J
a: In the moment the ball get shot out of the tube it has the potential energy in the feather gone over to kinetic energy in the ball. The ball therefore has kinetic energy.
Ep = 1/2mv^2 = 0.230J
We solve this and tries to find the speed (v): v = sqrt(2Ek / m) = sqrt((2*0.230 / (25.0 * 10^-3)) = 4.29m/s
b: If we put the feather cannon vertical, the ball gets potential energy on its way out of the cannon /tube. The potential energy in the feather goes therefore over to both potential energy and kinetic energy in the ball.
Ep, feather = Ep, ball + Ek
1/2kx^2 =mgh +1/2mv^2
Because we are looking at the speed at the exsact moment when it just passes the tube we set the height h = x.
1/2kx^2 =mgx +1/2mv^2
Vi are trying to find the speed so we solve it with the sake of v.
v= sqrt((kx^2-2mgx)/(m))
= sqrt((29.5*0.125^2-2*25.0*10^(-3)*9.81*0.125)/(25.0*10^(-3)) = 4.00 m/s
What I don't understand about this example is the kinetic and potential energy thingy. Exactly what it is (this may sound a bit noob is, but I failed physics don't laugh please).
And in a: it also says this: "In the moment the ball gets shot out of the tube, it has the potentioal energy from the feather gone to kinetic energy to the ball. And therefore the ball has kinetic energy". I don't understand how the ball got potential energy from the feather in the moment it get shot out horizonatally.. I also don't understand how the energy from the feather (the potential energy) can go to kinetic energy in the ball.)
in b: it says "If we put the feather cannon vertical, the ball get potential energy when its getting out of the cannon. The potential energy in the feather goes therefore over to potential and kinetic energy in the ball" (this is what I don't understand.. like how can potential energy in the feather go to potential and kinetic energy in the ball??)
Ohh by the way I haven't really studied english physics so some of the terms are very hard in english not sure if I translated kinetic and potential energy correctly.
Sincerly,
Failedphysics.
A feather or spring or whatever it's called got the stiffness 29.5 N/m. It is lying inside a horizontal tube (feather cannon). The feather /spring is connected in the very end of the tube. The feather / pring is as "long" as the tube if it's not being pressured longer into the tube or stretched out. We press the feather / spring 12.5cm in and then locks it there. Then we unlocks it and a ball gets shoot out.
(From now on I will call the spring or feather or whatever it is a feather, not sure if it's correctly used)
a) How big is the speed of the feather when it passes the end of the tube?
b) What will be the speed of the ball if we put the tube vertical and shoot the ball up?
Solution: When the feather is tense, it has potential energy: Ep = 1/2kx^2 = 1/2 * 29.5 N /m * (0.125m)^2 = 0.230 J
a: In the moment the ball get shot out of the tube it has the potential energy in the feather gone over to kinetic energy in the ball. The ball therefore has kinetic energy.
Ep = 1/2mv^2 = 0.230J
We solve this and tries to find the speed (v): v = sqrt(2Ek / m) = sqrt((2*0.230 / (25.0 * 10^-3)) = 4.29m/s
b: If we put the feather cannon vertical, the ball gets potential energy on its way out of the cannon /tube. The potential energy in the feather goes therefore over to both potential energy and kinetic energy in the ball.
Ep, feather = Ep, ball + Ek
1/2kx^2 =mgh +1/2mv^2
Because we are looking at the speed at the exsact moment when it just passes the tube we set the height h = x.
1/2kx^2 =mgx +1/2mv^2
Vi are trying to find the speed so we solve it with the sake of v.
v= sqrt((kx^2-2mgx)/(m))
= sqrt((29.5*0.125^2-2*25.0*10^(-3)*9.81*0.125)/(25.0*10^(-3)) = 4.00 m/s
What I don't understand about this example is the kinetic and potential energy thingy. Exactly what it is (this may sound a bit noob is, but I failed physics don't laugh please).
And in a: it also says this: "In the moment the ball gets shot out of the tube, it has the potentioal energy from the feather gone to kinetic energy to the ball. And therefore the ball has kinetic energy". I don't understand how the ball got potential energy from the feather in the moment it get shot out horizonatally.. I also don't understand how the energy from the feather (the potential energy) can go to kinetic energy in the ball.)
in b: it says "If we put the feather cannon vertical, the ball get potential energy when its getting out of the cannon. The potential energy in the feather goes therefore over to potential and kinetic energy in the ball" (this is what I don't understand.. like how can potential energy in the feather go to potential and kinetic energy in the ball??)
Ohh by the way I haven't really studied english physics so some of the terms are very hard in english not sure if I translated kinetic and potential energy correctly.
Sincerly,
Failedphysics.