Puck sliding accross ice W/ friction and incline

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a puck sliding up an incline with friction. The original poster describes the initial conditions, including the puck's weight, launch speed, and the incline's angle and friction coefficient, and seeks to determine various forces and distances related to the puck's motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster calculates the puck's acceleration and net force during its initial motion, then attempts to analyze the forces acting on the puck as it moves up the incline, including gravitational and frictional forces. Participants question the correctness of the gravitational force components and suggest alternative methods for finding the components of the gravitational force acting along the incline.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on the calculations and questioning assumptions about the forces involved. There is a focus on verifying the calculations related to the gravitational force components and the implications for the puck's motion.

Contextual Notes

There is a noted uncertainty regarding the accuracy of the gravitational force components and the overall setup of the problem, particularly in relation to the forces acting on the puck as it moves up the incline.

Evolution17
Messages
5
Reaction score
0
A puck, weighing 75KG(yes, 75kg) is launched from 0-20m/s in 4.4s. Determine the unbalanced force and acceleration.



Fnet=ma A=v/t



This PART was easy, A=20/4.4, A=4.55m/s^2. Fnet=75(4.55) Fnet=341.25N

Its the next part that gets tricky...

The puck, traveling at a constant 20m/s approaches a 15deg incline hill with a frictional mu of 0.100. Determine the force of friction, how far up the hill it will travel before coming to a complete stop, acceleration and the applied force.

So what I did was first find the Fg which is 735N
Then I split it up into X and Y components, for the Y component I ended up with 710N because COS15(735N)=710N. This should also be the normal force (I'm just not sure if I found the right component/value for Fg) so then Ff=Fn(mu) Ff=710(0.100) Ff=71N
Since the Ff is 71N, and there is NO applied force pushing it UP the incline, I did 71/75=-0.95M/s^2 and have that as my acceleration value. Then I subbed it into the V2^2=V1^2+2ad, I ended up subbing in 0=(20)^2+2(-0.95)D 400=1.9D D=210m. Therefore distance before stopping is 210M. It seems off to me so any confirmation/help would be greatly appreciated.
 
Physics news on Phys.org
You forgot another force acting on the puck along the incline besides friction.
 
Ah you're right, the force of gravity in the X direction. Would you mind checking if the value of 710N for the Y direction is 100% correct? And you would use Pyth theorem to solve for the other, so it would be 710^2+b^2=735^2 right?
 
Evolution17 said:
Ah you're right, the force of gravity in the X direction. Would you mind checking if the value of 710N for the Y direction is 100% correct? And you would use Pyth theorem to solve for the other, so it would be 710^2+b^2=735^2 right?

Yes, using the cos function is correct in finding the component of the gravity force perpendicular to the incline . You can use Pythagoras to find the comp of the gravity force parallel to the incline, but it might be easier to use the sin function.
 

Similar threads

Replies
2
Views
790
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
9
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
19K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K