How Does Air Force Affect Kinetic Energy in a Fancart?

Click For Summary
SUMMARY

The discussion focuses on calculating the change in momentum and kinetic energy of a fancart with a mass of 0.8 kg subjected to a constant force of <-0.4, 0, 0> N for 1.5 seconds. The change in momentum was determined to be <-0.6, 0, 0> kg·m/s. For kinetic energy, the user initially calculated values of 0.196 J and 0.225 J using the equations K = 1/2mv² and K = p²/2m, respectively, but did not successfully find the change in kinetic energy, which requires calculating the kinetic energy before and after the force is applied.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Familiarity with the concepts of momentum and kinetic energy
  • Ability to apply the equations K = 1/2mv² and K = p²/2m
  • Basic knowledge of vector notation in physics
NEXT STEPS
  • Calculate the initial kinetic energy of the fancart using K = 1/2mv²
  • Determine the final velocity of the fancart after the force is applied
  • Calculate the final kinetic energy using K = 1/2mv² after the force application
  • Find the change in kinetic energy by subtracting the initial kinetic energy from the final kinetic energy
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of momentum and energy calculations in real-world applications.

magma_saber
Messages
73
Reaction score
0

Homework Statement


A fancart of mass 0.8 kg initially has a velocity of < 0.7, 0, 0 > m/s. Then the fan is turned on, and the air exerts a constant force of < -0.4, 0, 0 > N on the cart for 1.5 seconds.

What is the change in momentum of the fancart over this 1.5 second interval?

What is the change in kinetic energy of the fancart over this 1.5 second interval?


Homework Equations


K = 1/2mv2
K = p2/2m

The Attempt at a Solution


I got the change in momentum. It's <-.6,0,0>
I can't find the change in kinetic energy.
I tried K = 1/2mv2 and got 0.196
I also tried p2/2m and got 0.225
 
Physics news on Phys.org
You must find the KE before and after. Take the difference to get the change.
 

Similar threads

Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
8
Views
2K
  • · Replies 58 ·
2
Replies
58
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 39 ·
2
Replies
39
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K