GRADE 12: Energy, Springs and maybe momentum/collisions?

In summary: Uniform_accelerationIn summary, the conversation discusses using equations for 2D trajectory and conservation of mechanical energy to calculate the distance a spring of mass "m" should be pulled back to land in a target that is "dx" distance away and "dy" meters tall. The spring has a constant of "k" and is launched from a launcher elevated at an angle of "θ" degrees. The classical range equation for 2D trajectory requires the initial and final elevations to both be on the ground, but in this problem there is an initial and final height. Therefore, the SUVAT equations for constant acceleration must be used to solve for the distance.
  • #1
michael simone
3
0

Homework Statement


A spring of mass “m” is put on a launcher elevated at “θ” degrees above the horizontal. It is pull back a distance “x” and launched at a target that must travel vertically “dy“ and a horizontal distance of “dx”. The spring has a constant of “k”.Derive an equation (or series of equations) given m, θ, dy ,dx, k so you can calculate a value of x.Example The target is 3.5 m away and the launder is elevated at 30o and it is 1.2 m high, the mass of the spring is 4.5 g, what distance should the spring be pulled back to land in a target that is 0.40 m tall if k = 18 N/m?

Homework Equations


PTi= PTf
Ei=Ef
ENERGY
KINETIC (Ek)
Ek=1/2 MV^2
SPRINGS(elastic Ee)
Ee=1/2 KX^2
E1=E2 IE
Potential (Eg)
Eg= mgh or mg(delta y)
vx= vicos (theta)
Projectile motion and Energy[/B]

The Attempt at a Solution

 
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  • #2
Check your textbook for 1) 2D trajectory equations and 2) the conservation of mechanical energy. You will need both.
 
  • #3
Dr Dr news said:
Check your textbook for 1) 2D trajectory equations and 2) the conservation of mechanical energy. You will need both.
ENERGY
KINETIC (Ek)
1/2 MV^2
SPRINGS(elastic Ee)
1/2 KX^2
E1=E2 IE
Potential (Eg)
mgh or mg(delta y)
vx= vicos (theta)
 
  • #4
The classical range equation is based on the initial elevation and the final elevation both being on the ground. In your problem you have an initial height as well as a final height which means you need to carry them along in your trajectory analysis.
 
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Likes michael simone
  • #5
thanks for that tip, but how do i slove this when v is not provided.
 
  • #6

Related to GRADE 12: Energy, Springs and maybe momentum/collisions?

1. What is potential energy and how is it related to springs?

Potential energy is the energy that an object possesses due to its position or state. In the context of springs, potential energy is stored in the deformation of the spring when it is compressed or stretched. This potential energy can be converted into kinetic energy when the spring is released and returns to its original shape.

2. How is the potential energy of a spring calculated?

The potential energy of a spring can be calculated using the formula PE = 1/2kx^2, where PE is potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position. This formula assumes that the spring is an ideal spring with no mass and follows Hooke's Law, which states that the force applied to a spring is directly proportional to its displacement.

3. Can potential energy be negative in the context of springs?

Yes, potential energy can be negative in the context of springs. This occurs when the spring is compressed and its displacement is negative. Negative potential energy simply indicates that the spring has the potential to release energy and return to its equilibrium position.

4. What is momentum and how is it conserved in collisions?

Momentum is the product of an object's mass and velocity, and it is a measure of its motion. In collisions, momentum is conserved, meaning that the total momentum of the objects before the collision is equal to the total momentum after the collision. This is due to the law of conservation of momentum, which states that in a closed system, the total momentum remains constant.

5. How does the mass and velocity of objects affect the outcome of a collision?

The mass and velocity of objects have a significant impact on the outcome of a collision. In an elastic collision, where there is no loss of kinetic energy, the objects with greater mass and velocity will have a greater impact force and will cause the other object to move with a greater velocity. In an inelastic collision, where there is a loss of kinetic energy, the objects with greater mass and velocity will have a greater decrease in velocity after the collision.

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