# Hockey puck friction problem

#### evelynn33

Hello, i'm new and need help! Here's the deal...

You have a hockey puck that's mass 2.5lbs. You shoot it up a ten degree incline at 8ft/sec. The friction factor on the ramp is .08. You are trying to reach a hole that is 10 feet away. Does the puck make it?

In trying this problem for the last three days, I have had several different answers, but none make much sense.

Do you have to convert the 2.5lbs to mass in order to get the answer? I've tried changing it (2.5lbs) to force (N), but then get stuck after I add up all the forces acting on the puck and make them equal to mass X acceleration.

My prof. said you have to convert lbs. to force which I did, but I still think you need to get the mass.

Any help or suggestions would be greatly appreciated.
Evelynn

As for work I don't have the exact numbers right in front of me, but I'll do my best.

First, 2.5lbs = -11.12 N that's the force of gravity (I'm assuming)
Then, cos 10 * 11.12N to get the Normal force
then sin 10 * 11.12N to get the x component of gravity force
then .08 * Normal force to get the friction force (negative)

Then you add the Friction force & the X component of gravity and set that = to mass * acceleration. (Am I right so far?)

Here's where i get stuck. I've tried using the 2.5 as the mass, and I've tried converting the 2.5 (as a weight) to mass by dividing by 32 ft/sec.
Anyway, after you get acceleration you can use the kinmatics equations to get time and final position...which you want to know if it is around 10 ft. The final answers I've gotten are either 13.something, or 1.something. And I've tried manipulating the numbers through conversions of feet to meters, but just get even more confused.

P.S. Do I need to change all my units to m/s?

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#### Zimm

Well you don't have to convert anything if you don't want, using english units can be fun! You just need to understand that pounds is a weight or force the way newtons is and that slugs are the english unit of mass, which when multiplied by the acceleration of gravity in english units, 32 ft/s^2, gives pounds.

So what you want to do is draw a picture with triangles to show the ramp and the force normal. And you'll see that the force of gravity Fg=2.5lbs*sin(10)=.434lbs which is down the ramp. And then you look at the force normal Fn=2.5lbs*cos(10)=2.46lbs which you use with the coefficient of friction u=.08=Ff/Fn to get the force of friction Ff=u*Fn=.08*2.46lbs=.197lbs down the ramp

Now, you have the force on the puck down the ramp, F=Fg+Ff=.434+.197=.631lbs and so you have effectively changed your coordinate system because you're just talking about up and down the ramp, parallel to the ramp. But now you need the acceleration on the puck so use Newton's second law variation F=ma -> a=F/m and m=weight/gravity=2.5lbs/(32ft/s^2)=.078 slugs then you do a=F/m=.631lbs/.078slugs=8.08ft/s^2 down the ramp, and it is ft/s^2, if the units are unfamiliar then think about the metric equivalent, if you took Newtons/kg you would have m/s^2

So now you need the equation to relate velocity to acceleration and distance, v^2=2ad, I always integrate to get this and you take F=ma -> ma=ma, and cancel the mass to get a=a, and on the one side change "a" to -8.08ft/s^2 (negative because it is back down the ramp, the negative is very important) so -8.08ft/s^2=a and a=dv/dt, distance over time, but also a=(dv/dx)*(dx/dt)=v*dv/dx so you have -8.08ft/s^2=v*dv/dx or (-8.08ft/s^2)*dx=v*dv then you integrate both sides over appropriate limitsm dx goes from 0 to 10 (remember that I said we basically changed coordinates so it's like the pucks on the floor now) and the velocity goes from 8ft/s to vf (v-final) and this gives you a*d=(vf^2)/2 in case you're doing it algebraically, so the evaluated integral is (-8.08ft/s^2)*10ft=((vf^2)/2) - ((8ft/s)^2) /2

Now when you solve that you see why the negative is important, because you get vf=sqrt( ((-8.08ft/s^2)*10) + ((8ft/s)^2) /2 ) and you see you have the square root of a negative number and this tells you that the puck did not make it.

Now if you want to see how far the puck will go you change the integral limits so the dx goes from 0 to xf (x-fnial) and the dv goes from 8ft/s to 0ft/s and this gives you (-8.08ft/s^2)*xf=- ((8ft/s)^2)/2 or xf=(-32(ft/s)^2) / (-8.08ft/s^2) = 3.96 ft so it's nowhere near 10 ft.

#### krab

As Zimm said, the friction force along the plane is

W&mu; cos&theta;

(&mu; is the coefficient of friction, W is the weight of the puck)
The component of gravity along the plane is

W sin&theta;

Add these together and equate to ma, which is the same as (W/g)a. (g is the acceleration of gravity.) Notice that the weight W cancels. This is common in problems of this type. So you needn't have worries about how to convert the 2.5lbs to other units.

So

a=g(&mu;cos&theta; + sin&theta;)= 8.08 ft/s/s

Now use v^2=2ad for d the distance the puck goes when initial speed is v.

d=v^2/(2a)= 64/(2*8.08) ft = 3.96 ft

#### evelynn33

Thank you sooo much!!

I guess I was on the right track, but it seemed to me the answer should have been closer to 10 ft, particularly because of the initial velocity. I may get this Physics stuff yet!

Thank you again
evelynn

#### krab

I dislike teaching physics by advising "Just use this formula and you get that answer", because in that approach, no physics is learned, and the student is left wondering: "How do I know to use that formula and not another?" I was guilty of that in my last post. So here's a better way.

Notice the deceleration of the puck is 8.08m/s/s, and that the initial speed is 8m/s. That means that the puck will lose all of its speed in just under 1 second. If it had travelled at constant speed, it would travel 8 metres in that 1 second. But we know the speed is linearly dropping with time, starting 8m/s, ending 0m/s, so we cut the 8 m guess in half.

To formalize: we find the time taken, t=v/a, and note the distance travelled as d=vt/2=(v/2)(v/a)=v^2/(2a).

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