Puck sliding accross ice W/ friction and incline

[Evolution17]

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.

 

[PhanthomJay]

You forgot another force acting on the puck along the incline besides friction.

 

[Evolution17]

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]

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.

 

[Nickie]

I would first like to commend you on your efforts in solving this problem. It is clear that you have a good understanding of the concepts involved and have applied them correctly. However, there are a few points that I would like to address to ensure accuracy in your calculations.

Firstly, the value of the puck's weight is given as 75 kg, not 75 N. This means that the puck's weight is actually 735 N, not 75 N as you have used in your calculations. This may have led to some discrepancies in your final calculations.

Secondly, in order to find the normal force (Fn), we need to consider the forces acting in the y-direction (perpendicular to the incline). These forces are the puck's weight (735 N) and the normal force (Fn). As you correctly calculated, the y-component of the puck's weight is 710 N. Therefore, the normal force would also be 710 N, not 735 N as you have used in your calculations. This will also affect the value of the frictional force (Ff), which you have correctly calculated as 71 N.

Next, in order to find the acceleration of the puck, we need to consider all the forces acting on it in the x-direction (parallel to the incline). These forces are the frictional force (Ff) and the applied force (Fa), which in this case is zero. Therefore, the net force in the x-direction (Fnetx) would be equal to the frictional force (Ff). Using Newton's second law (Fnet=ma), we can calculate the acceleration as a=Ff/m. Substituting the values, we get a=71 N/75 kg=0.95 m/s^2. This is the correct value for the acceleration, not -0.95 m/s^2 as you have used in your calculations.

Finally, to find the distance traveled by the puck up the incline before coming to a complete stop, we can use the equation V^2=V0^2+2ad, where V is the final velocity (which is zero), V0 is the initial velocity (which is 20 m/s), a is the acceleration (which we have calculated as 0.95 m/s^2), and d is the distance traveled. Substituting the values, we get 0=(20)^2+2(0.