Kinetic Energy Gain: Why Is There a Difference?

  • Thread starter Thread starter WiFO215
  • Start date Start date
  • Tags Tags
    Energy
AI Thread Summary
When a car moving at speed u throws a ball at speed v relative to itself, the kinetic energy gained is perceived differently by the car's occupant and an observer on the ground. The occupant measures a gain of mv²/2, while the ground observer calculates a change of m[v² + 2uv]/2, highlighting the discrepancy in energy perception due to different reference frames. This difference arises because the kinetic energy of both the car and Earth changes, affecting the total energy calculation. The discussion references a previous thread on "DDWFTTW" propulsion, emphasizing the importance of agreeing on a reference frame when discussing energy sources. The example of a roller skater throwing an object illustrates that the energy of the thrown object includes both the skater's effort and their momentum.
WiFO215
Messages
416
Reaction score
1
Say a car is moving with speed u. The guy in the car chucks a ball with speed v relative to himself. The gain in KE,as recorded by him will be mv2/2. For an observer on the ground, i.e. non-moving frame, the change in kinetic energy is [(v+u)2 - u2]m/2 = m[v2 + 2uv]/2. Why is there a difference in energy gained? I am not able to put my finger on it.
 
Physics news on Phys.org
Because there is a difference in how much the kinetic energy of the Earth and car has changed. (Do the math, imagine what happens when a roller-skater throws a brick.)

There was a long thread here previously about "DDWFTTW" propulsion, and your topic is the reason why it never made much sense there to argue (without agreeing on a reference frame) whether the cart's power was "coming from" either the air at the propeller or the ground at the wheels.
 
Last edited:
I'm going to use the example Cesiumfrog used.

Think of a roller skater throwing a ball or a brick, the amount of energy the object thrown has is not just the energy that the roller skater threw the object, but also the momentum energy the roller skater had while he threw it
 
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Dear all, in an encounter of an infamous claim by Gerlich and Tscheuschner that the Greenhouse effect is inconsistent with the 2nd law of thermodynamics I came to a simple thought experiment which I wanted to share with you to check my understanding and brush up my knowledge. The thought experiment I tried to calculate through is as follows. I have a sphere (1) with radius ##r##, acting like a black body at a temperature of exactly ##T_1 = 500 K##. With Stefan-Boltzmann you can calculate...
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (First part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8 and stuck at some statements. It's little bit confused. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. Solution : The surface bound charge on the ##xy## plane is of opposite sign to ##q##, so the force will be...
Back
Top