2 Things: Work-Energy theorem and parabolic trajectories.

In summary, the work-energy theorem states that the total work done on an object is the sum of the work done on each of its parts.
  • #1
misogynisticfeminist
370
0
Hi,

When reading through the work-energy thread, i just got reminded of something bugging me for quite a while. I don't think this is mentioned in the work-energy thread, firstly, how can we show that,

[tex] \int f \bullet ds = \Delta E [/tex]

also,

I've heard that there's a way to show that when only one force acts on a particle or something, the particle follows a parabolic trajectory. Is there a way to show this?

thanks...
 
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  • #2
If there is a constant (non-zero) force acting on something, it follows a parabolic trajectory. That's because, taking the x-axis perpendicular to the constant force, integrating a constant twice (from [itex]\frac{d^2y}{dt^2}= \frac{F}{m}[/itex] to [itex]\frac{dy}{dt}= \frac{F}{m}t+ v_0[/itex] to [itex]y= \fra{F}{2m}t^2+ v_0t+ y_0[/tex]) we get a quadratic in t while x, with 0 acceleration, is linear in t.

(There's always "only one force" on something- the net force.)
 
  • #3
HallsofIvy said:
If there is a constant (non-zero) force acting on something, it follows a parabolic trajectory. That's because, taking the x-axis perpendicular to the constant force, integrating a constant twice (from [itex]\frac{d^2y}{dt^2}= \frac{F}{m}[/itex] to [itex]\frac{dy}{dt}= \frac{F}{m}t+ v_0[/itex] to [itex]y= \fra{F}{2m}t^2+ v_0t+ y_0[/tex]) we get a quadratic in t while x, with 0 acceleration, is linear in t.

(There's always "only one force" on something- the net force.)

ohhhh ok, that helped alot, thanks..

Anyone with the work question?
 
  • #4
You can't derive the work formula, as it is a definition. You can, however justify why the definition is a good one with a little bit of logical reasoning.

Claude.
 
  • #5
But you can certainly derive the Work-Energy theorem. Taking the simplest case, where the direction of the net force remains fixed:
[tex] \int f \bullet ds = \int m a \bullet ds = m \int \frac{dv_s}{dt} ds = m \int v_s \ dv_s = \Delta (1/2 m v_s^2) [/tex]
 
  • #6
oh ok thanks, that's what I'm looking for.

: )
 

1. What is the Work-Energy theorem and how does it relate to physics?

The Work-Energy theorem states that the work done on an object is equal to the change in its kinetic energy. In other words, the amount of work done on an object is directly proportional to the change in its velocity. This theorem is fundamental to the study of physics as it helps us understand the relationship between work and energy.

2. How is the Work-Energy theorem applied in real-life situations?

The Work-Energy theorem is applied in many real-life situations, such as calculating the energy needed to launch a rocket into space or determining the speed of a moving car. It is also used in designing machines and structures, as engineers need to consider the work and energy involved in their operation.

3. Can you explain the concept of parabolic trajectories?

A parabolic trajectory is the path that an object takes when it is thrown or launched near the surface of the Earth. It follows a curved path that resembles a parabola, hence the name. This trajectory is a result of the combination of the object's horizontal and vertical velocities and the force of gravity acting on it.

4. How do the Work-Energy theorem and parabolic trajectories intersect?

The Work-Energy theorem is often used to analyze the motion of objects in a parabolic trajectory. By applying this theorem, we can calculate the work done on the object and its change in kinetic energy at any point along its path. This helps us understand the factors that affect the object's trajectory, such as its initial speed and angle of launch.

5. Are there any real-life examples of parabolic trajectories?

Yes, there are many real-life examples of parabolic trajectories, such as a thrown ball, a basketball shot, or a fireworks display. These objects follow a parabolic path due to the force of gravity acting on them. Understanding parabolic trajectories is also crucial in fields such as astronomy and missile engineering.

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