Net Forces & Friction

  • #1

Homework Statement



A resting 164 g hockey puck is hit with a force of 15.3N. The frictional force slowing the puck down is 0.75N.

a) Find the net forces acting on the puck. Remember to consider both horizontal and vertical forces
b) If the puck leaves the stick with a velocity of 45m/s how far will the puck travel in 3.0s?

m= 0.164kg
F[itex]_{applied}[/itex]= 15.3N [forward]
F[itex]_{f}[/itex]= 0.75N [backwards]

Homework Equations



For part b = a=2(d)/t²


The Attempt at a Solution


a- 15.3[forward] - 0.75 [backwards]
= 14.55N [forward]
The puck would have a net force of 14.55N.
(it doesn't seem right because the questions is telling me to remember the vertical forces as well.)

b- a=2(d)/t²
45m/s = 2(d)/3²
45m/s x 9 = 2(d)/9 x 9
405 m = 18(d)
22.5 m = d
Would that be correct?? It just seems off.
 

Answers and Replies

  • #2
Doc Al
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The Attempt at a Solution


a- 15.3[forward] - 0.75 [backwards]
= 14.55N [forward]
The puck would have a net force of 14.55N.
(it doesn't seem right because the questions is telling me to remember the vertical forces as well.)
The question is a bit ambiguous. If they meant the net force while it's being hit, then you're right. (And I also don't know why they mention vertical forces.)

b- a=2(d)/t²
45m/s = 2(d)/3²
45m/s x 9 = 2(d)/9 x 9
405 m = 18(d)
22.5 m = d
Would that be correct?? It just seems off.
Several problems. First, what's the acceleration once the puck leaves the stick? Second, be sure to use a more general formula that includes the initial velocity.
 
  • #3
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One thing that you can do is create questions that it isn't asking. Question that might help check your answer. For example, you know that the puck is accelerating from 45 m/s to some unknown final velocity after 3 sec. If you assume it isn't accelerating then you can figure out how far it will travel in 3 sec knowing its velocity is constant (zero acceleration by assumption). This is an easy equation

v=d/t
d=v*t=(45m/s)*(3sec)=135m

This means the puck will travel at least 135 meter with no acceleration. If you say now that it is accelerating (as the problem states), then you know the answer for the distance should be larger than this 135 meters.

In regards to the part a of this question. Most physics problems try to get you in the habit of drawing a "Free-Body Diagram." This seem unnecessary since you assume that the puck isn't doing anything interesting in the up and downward direction but it is good practice to check this anyway. Summing the forces is much easier once all the forces have been considered on the free body diagram.
 
  • #4
Several problems. First, what's the acceleration once the puck leaves the stick? Second, be sure to use a more general formula that includes the initial velocity.
Would the formula for finding the acceleration once the puck leaves the stick be:
a=v/t

a=45m/s / 3.0s)
a=15m/s²

so then would it be:

a=2(d)/t²
15=2(d)/9
15x9 = 2(d)x9
135 = 18d
135/18 = 18d/18
7.5m?

So in this case the puck would travel 7.5m.
 
  • #5
Doc Al
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Would the formula for finding the acceleration once the puck leaves the stick be:
a=v/t

a=45m/s / 3.0s)
a=15m/s²
No. With that formula, you assumed that the puck goes from 45m/s to 0 in 3 seconds. But no one said it comes to rest.

Instead, use Newton's 2nd law to find the acceleration. Once you have the correct acceleration, you can find the speed at the end of those 3 seconds.
 
  • #6
No. With that formula, you assumed that the puck goes from 45m/s to 0 in 3 seconds. But no one said it comes to rest.

Instead, use Newton's 2nd law to find the acceleration. Once you have the correct acceleration, you can find the speed at the end of those 3 seconds.
f=ma
14.55N = (0.165g)(a)
14.55/0.165g = (0.165g)(a)/0.165g)|
88.19m/s² = a

a=2(d)/t²
88.19m/s² = 2(d)/3²
88.19m/s²x9s = 2(d)/9 x 9
793.71m = 18d
793.71m/18 = d
44.095m

44.1m

Is that the correct way to do it??
Did I ever mention how much I dislike the hundreds of different formulas physics actually has! (ps thanks for the help)
 
  • #7
Doc Al
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f=ma
14.55N = (0.165g)(a)
14.55/0.165g = (0.165g)(a)/0.165g)|
88.19m/s² = a
You have the wrong force. 14.55 N is the net force while the puck is being hit. But after the puck leaves the stick (moving at 45 m/s), the only force on it is the friction acting to slow it down.
 
  • #8
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Another clue when doing physics problems like this one comes from what is used from the problem to get the solution. The problems tells you that the puck travels with an initial velocity of 45m/s. However in your calculations you haven't used this initial velocity at all. This would imply that the distance the puck travels in 3 sec is independent of it's initial velocity, which is nonsensical. I would try to review you starting formula for part b and see if there is a way to include this initial velocity.

Also, remember the check of answer as I mentioned before. If you achieve an answer that is less than 135 m (the distance achieved with zero acceleration), the acceleration would have to be negative, that is the puck is decelerating, which you know is not the case.

I hope this helps!
 
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  • #9
You have the wrong force. 14.55 N is the net force while the puck is being hit. But after the puck leaves the stick (moving at 45 m/s), the only force on it is the friction acting to slow it down.
Thanks! I got the answer.. thanks for everything. Out of 30 odd questions this and another one plus another is the only ones I had trouble with.
 
  • #10
"You have the wrong force. 14.55 N is the net force while the puck is being hit. But after the puck leaves the stick (moving at 45 m/s), the only force on it is the friction acting to slow it down."

Hey everyone I'm a first time user and had the same problem with a similar question just different numbers. I've been working along with this example and I can't figure out what the force is, so I can't figure out the acceleration and in turn can't solve the problem. If you aren't using 14.55 N, how do you incorporate the 45 m/s?
 
  • #11
Doc Al
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If you aren't using 14.55 N, how do you incorporate the 45 m/s?
You must use the correct force to get the correct acceleration. As I stated previously, after the puck leaves the stick the only force acting is friction. The 45 m/s is the initial speed of the puck as it leaves the stick. First find the acceleration.
 
  • #12
okay... in my University text book, it says

a= V^2 - Vo^2
------------
2 Delta X X in this case being time right?

a= 0 - (45 m/s)^2
--------------
2(3)

a = -225 m/s^2
I thought it couldn't be negative acceleration, that's where I'm having trouble.
 
  • #13
Doc Al
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okay... in my University text book, it says

a= V^2 - Vo^2
------------
2 Delta X X in this case being time right?
No. In that equation, Δx would be the distance traveled, not the time.

Find the acceleration using dynamics--Newton's 2nd law. Once you have the acceleration, then use kinematics to solve for the distance traveled in the given time.

And there's no problem in having a negative acceleration. Depending on your sign convention, it just means that the object is slowing down.
 

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