A question for anyone knowledgeable in physics
Moderator: Alyrium Denryle
After the force has been applied. At rest your velocity is zero.Wooh wrote:initial velocity after the force has been applied or at rest?
By the way, since you seem to be qualified on the subject, what is it about physics that is so hard for normal people to understand (no offense intended)?
data_link has resigned from the board after proving himself to be a relentless strawman-using asshole in this thread and being too much of a pussy to deal with the inevitable flames. Buh-bye.
Lol, I am not qualified at all, thanks for the sarcasm 
I knew that it was 0 at rest..which is why I asked. And I get...some of this, but I take physics next year and haven't really done anything like it before. I get what I can. And when you say initial velocity, etc, when you specify exactly what it is it makes it a little bit easier than time-dog + velocity-cat = parrot-piano, it makes no sense if you are an uninitiated person who doesn't know the lingo
I knew that it was 0 at rest..which is why I asked. And I get...some of this, but I take physics next year and haven't really done anything like it before. I get what I can. And when you say initial velocity, etc, when you specify exactly what it is it makes it a little bit easier than time-dog + velocity-cat = parrot-piano, it makes no sense if you are an uninitiated person who doesn't know the lingo
Well, first of all, it's not lingo, it's terminology .
Second, here are some terms you might like to know:
meter: standard unit of length
second: standard unit of time
kilogram: standard unit of mass (yes, the standard unit does have a prefix)
velocity: your speed in a given direction, measured in m/s (that's meters per second)
acceleration: rate of change in velocity. Measured in units of m/s^2 (that's meters per second per second or meters per second squared)
Force: A push or a pull. Defined as the acceleration that it can impart to a standard unit of mass. Measured in N (newtons).
Newton: 1 kg*m/s^2
Momentum: The product of an object's mass and velocity. Measured in kg*m/s
Impulse: A change in momentum. Measured in N*s (if you do the math, you find that that is the same unit as momentum)
Work: The product of a force and the distance over which it is applied. Measured in N*m
Energy: The ability to do work. Measured in J (joules). 1 J = 1 N*m.
Power: The rate of transfer of energy. Measured in W (watts). 1 W = 1 J/s
Some useful equations:
v = d/t
a = v/t
d = d0 + v0t + .5at^2
F = ma
E = Fd
E = .5mv^2
p (momentum) = mv
Impulse = Ft
That should get you through at least half the class.
.Second, here are some terms you might like to know:
meter: standard unit of length
second: standard unit of time
kilogram: standard unit of mass (yes, the standard unit does have a prefix)
velocity: your speed in a given direction, measured in m/s (that's meters per second)
acceleration: rate of change in velocity. Measured in units of m/s^2 (that's meters per second per second or meters per second squared)
Force: A push or a pull. Defined as the acceleration that it can impart to a standard unit of mass. Measured in N (newtons).
Newton: 1 kg*m/s^2
Momentum: The product of an object's mass and velocity. Measured in kg*m/s
Impulse: A change in momentum. Measured in N*s (if you do the math, you find that that is the same unit as momentum)
Work: The product of a force and the distance over which it is applied. Measured in N*m
Energy: The ability to do work. Measured in J (joules). 1 J = 1 N*m.
Power: The rate of transfer of energy. Measured in W (watts). 1 W = 1 J/s
Some useful equations:
v = d/t
a = v/t
d = d0 + v0t + .5at^2
F = ma
E = Fd
E = .5mv^2
p (momentum) = mv
Impulse = Ft
That should get you through at least half the class.
data_link has resigned from the board after proving himself to be a relentless strawman-using asshole in this thread and being too much of a pussy to deal with the inevitable flames. Buh-bye.
What values are you inputting?Wooh wrote:Also, I used your formula, however, it got infinitely large...it never peaked and decreased, as it should. Perhaps I need to change formulas at some point, enlighten me, please.
data_link has resigned from the board after proving himself to be a relentless strawman-using asshole in this thread and being too much of a pussy to deal with the inevitable flames. Buh-bye.
2 for the force
10 seconds for amount of time the force is applied, going to assume that the gravity is the same as earths, and the mass of the object is 2.
Inputting that gives me
h=vi - 0.5 * (-9.8 ) * t^ 2
well, above it was mentioned that vi = Ft(I am assuming that this t is how long the force is applied)/m
soo, i substitute and get
h=Ft/m- 0.5 * (-9.8 ) * t^ 2, substitute in values and get
h=10-.5*(-9.8 )*t^2
I am assuming here that the force we are dealing with is net force...if not, then the force would have to be greater than 2, right (mass * gravity)? Well, if that is the case changing in 3 for the results doesn't effect it, so that equation above is what I am using.
10 seconds for amount of time the force is applied, going to assume that the gravity is the same as earths, and the mass of the object is 2.
Inputting that gives me
h=vi - 0.5 * (-9.8 ) * t^ 2
well, above it was mentioned that vi = Ft(I am assuming that this t is how long the force is applied)/m
soo, i substitute and get
h=Ft/m- 0.5 * (-9.8 ) * t^ 2, substitute in values and get
h=10-.5*(-9.8 )*t^2
I am assuming here that the force we are dealing with is net force...if not, then the force would have to be greater than 2, right (mass * gravity)? Well, if that is the case changing in 3 for the results doesn't effect it, so that equation above is what I am using.
Oh, there's your problem... you're inputting a negative value for gravity. I designed this equation based on the assumption that you would be using a positive value.
data_link has resigned from the board after proving himself to be a relentless strawman-using asshole in this thread and being too much of a pussy to deal with the inevitable flames. Buh-bye.
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