Hooke's Law and distance

1. Nov 25, 2003

marshall4

How do i find the distance that a spring will go using the equation
E=0.5kx^2, k & x is given.

Do i make that equation equal to 0.5mv^2 and solve for v? What would i do with v after that?

2. Nov 25, 2003

E = &frac12;kx2 appears to be the equation for the elastic potential energy of the spring. The force on the spring is the first derivative with respect to extension of this, or

F = kx

Re-arranging,

x = F/k

If you place a mass, m, on the spring, the extension will then be

x = mg/k

Where g is the acceleration due to gravity, i.e. the numerator is the mass's weight.

3. Nov 25, 2003

marshall4

How do i find how far a spring will shoot from a fixed angle?
Is there an equation similar to E=0.5kx^2 but that has an angle(theta) in it?

4. Nov 25, 2003

FZ+

AD: I think you are going the extra mile. Energy considerations is indeed the easiest way to do it.

$$KE_{in} = \frac{1}{2}mv^2 = Max EPE = \frac{1}{2}kx^2$$

From an angle...? Depends on what you are modelling the spring as. ie. if it is free to rotate to fit the line of your pull.

It all depends on the specific situation, though.

5. Nov 25, 2003

Oh, I'm sorry. I misunderstood the question. I thought Marshall was asking how far a spring would extend after you put a mass on it. I see what he means now. He's asking how fast it will be able to shoot a projectile.

In that case, yes, the kinetic energy of the projectile will be equal to the stored potential energy of the spring once it is fired.

If you want to find out how far the projectile will go given the initial velocity once it is airborne and its angle of projection, you can find its vertical component of velocity and figure out how long it will be in the air for. You can then multiply this by its horizontal component of velocity and find the horizontal distance travelled.

Another way of getting the range is using the formula:

Range = (v02 sin 2&theta;)/g

Where v0 is the initial velocity and &theta; the angle of projection.