Puck sliding accross ice W/ friction and incline

AI Thread Summary
The discussion focuses on calculating the forces acting on a 75 kg puck sliding up a 15-degree incline with a friction coefficient of 0.100. The initial acceleration of the puck is determined to be 4.55 m/s², resulting in a net force of 341.25 N. When analyzing the incline, the gravitational force is split into components, with the normal force calculated as 710 N. The force of friction is found to be 71 N, leading to an acceleration of -0.95 m/s². The distance traveled up the incline before stopping is calculated to be 210 m, though participants discuss verifying the accuracy of the gravitational components used in the calculations.
Evolution17
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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.
 
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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.
 
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