Understanding derivation of kinetic energy from impulse?

In summary, the conversation discusses the relationship between impulse, work, and kinetic energy. It is mentioned that the area under a force-time plot represents the impulse imparted on an object, which is equivalent to the change in momentum. The concept of integrating with respect to velocity is also explored, leading to the understanding that work is equal to 1/2mv^2, which is the kinetic energy of the object. The speaker also realizes their mistake of trying to integrate with respect to time and gains a better understanding of the concept.
  • #1
richsmith
4
0
I have a question that annoys my basic understanding of kinetic energy.

I know if I have a force-time plot then the area under the curve is equivalent to the impulse imparted on an object (in units Newton-Seconds). I know that this is also equivalent to the change in momentum of the object i.e. Ft = delta mV

I know that i can get these values from the plot by simply intgrating the Force function wrt to time. Now I want to determine the kinetic energy involved in this event. I know that k.e. = 1/2mV^2 so I know that is simply the integral of the impulse w.r.t. to velocity

Now that is the part conceptually I don't really grasp. What does it really mean to integrate with respect to velocity? This means I am suddenly on a velocity :uhh: domain, not time, and I don't really understand this?

Thanks.
 
Physics news on Phys.org
  • #2
While impulse is ∫Fdt, work is ∫Fdx. Since force can be viewed as the rate of change of momentum, you have F = dp/dt = m dv/dt. Thus:
W = ∫Fdx = ∫m (dv/dt) * dx = m∫dv*(dx/dt) = m∫v dv = ½mv² (which is kinetic energy).

Make sense?
 
  • #3
After some head scratching yes it does, finaly.

I think my mistake has always been trying to integrate wrt time. I really had to step back and ask myself what i was after, work that is, which is not a time integral.

Manipulating dv/dt to dx/dt and then subbing in V now makes it algebraicly understandable.

Thanks alot.

Richard.
 
  • #4
I have a question that annoys my basic understanding of kinetic energy.

Similar to questions which annoys my basic understanding of electrostatics?

Why not just calculate the final velocity of the body and then calculate the K.E.

Take as a caveat that the momentum of a body might not be directly proportional to K.E possesed by it.
 

1. What is kinetic energy?

Kinetic energy is the energy that an object possesses due to its motion. It is a scalar quantity that is dependent on the mass and speed of an object.

2. How is kinetic energy related to impulse?

Kinetic energy is derived from impulse by using the equation KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity. Impulse is defined as the change in momentum of an object, and momentum is equal to mass times velocity. Therefore, by combining these equations, we can derive the relationship between kinetic energy and impulse.

3. What is the formula for calculating impulse?

The formula for impulse is J = F * ∆t, where J is impulse, F is the average force applied to an object, and ∆t is the time interval over which the force is applied. This formula can also be written as J = ∫ F dt, where the integral represents the total area under the force-time curve.

4. How does the magnitude of impulse affect the change in kinetic energy?

The magnitude of impulse is directly proportional to the change in kinetic energy. This means that the greater the impulse applied to an object, the greater the change in its kinetic energy will be. This is because impulse is what causes a change in an object's momentum, which in turn affects its kinetic energy.

5. Can the derivation of kinetic energy from impulse be applied to all types of motion?

Yes, the derivation of kinetic energy from impulse can be applied to all types of motion, as long as the motion is linear and the forces involved are constant. This is because the equations for impulse and kinetic energy are based on the principles of conservation of momentum and energy, which apply to all types of motion.

Similar threads

  • Mechanics
Replies
11
Views
929
Replies
16
Views
1K
Replies
21
Views
4K
Replies
5
Views
787
Replies
7
Views
817
Replies
2
Views
1K
  • Mechanics
Replies
12
Views
1K
Replies
1
Views
487
Replies
4
Views
934
Replies
14
Views
2K
Back
Top