# Puck sliding accross ice W/ friction and incline

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.

Related Introductory Physics Homework Help News on Phys.org
PhanthomJay
Homework Helper
Gold Member
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?

PhanthomJay