Hocky puck vel/displacement/acc

  • Thread starter Thread starter DarkTangent
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on calculating the acceleration, final velocity, and displacement of a hockey puck on a horizontal air table, subjected to a force of 1.00 N. The puck has a mass of 0.250 kg. Using Newton's second law (F=ma), the acceleration is determined to be 4 m/s². The participant also discusses the correct application of the equation for acceleration, confirming their understanding of the physics involved.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Basic knowledge of kinematics equations
  • Familiarity with concepts of mass and force
  • Ability to perform calculations involving acceleration and displacement
NEXT STEPS
  • Learn how to calculate final velocity using kinematic equations
  • Explore the concept of displacement in uniform acceleration scenarios
  • Study the effects of friction on motion in physics
  • Investigate the design and functionality of air tables in physics experiments
USEFUL FOR

Physics students, educators, and anyone interested in understanding motion dynamics and the application of Newton's laws in practical scenarios.

DarkTangent
Messages
3
Reaction score
0
1. By pumping air up through thousands of tiny holes we can make a horizontal air table that will support several pucks so that they can move around on cushions of air with very little friction. Suppose that one such puck initially at rest, having a mass of 0.250 kg, is pushed along by a 1.00 N force for 2.00 s. Determine its acceleration and the final velocity at the end of 2.00 s. What will be the displacement?



2. Well, of all the equations that I thought I should use, I thought it would be Acceleration=(Velocity final-Velocity initial)/(time)



3. I've made several attempts at a solution but I'm not sure if I'm using the right equation, someone mind pushing me in the right path?
 
Physics news on Phys.org
Never mind. I figured out that f=ma and I know the force and the mass. No biggy :)
The rest will be easy to figure out, thanks :)
Though, it would be (1 kg m/s^2)/ (.250 kg) = 4 m/s^2, correct?
 

Similar threads

Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 56 ·
2
Replies
56
Views
5K
  • · Replies 6 ·
Replies
6
Views
7K
  • · Replies 11 ·
Replies
11
Views
2K