What is the acceleration experienced by the puck?

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Homework Help Overview

The discussion revolves around a physics problem involving the acceleration of a hockey puck sliding over ice, initially moving at a speed of 5.5 m/s and coming to a stop over a distance of 15 m due to friction. Participants are exploring the calculation of acceleration in this context.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are discussing the use of kinematic equations to find acceleration, questioning the correctness of initial calculations, and clarifying the relationship between displacement, initial and final velocities, and acceleration.

Discussion Status

The conversation is active, with participants sharing their attempts and corrections. Some guidance has been offered regarding the use of appropriate equations and the need for algebraic manipulation. There is an ongoing exploration of the assumptions regarding uniform acceleration.

Contextual Notes

Participants have noted the importance of distinguishing between initial and final velocities and the implications of negative acceleration due to deceleration. There is also mention of a helpful external resource provided for further clarification.

petal5
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I've been attempting the following question:

A 120gram hockey puck sent sliding over ice with an initial speed of 5.5m/s is stopped in 15m by the frictional force on it from the ice.What is the acceleration experienced by the puck?

I'm getting an answer of 2.0157m/s^2.I'm wondering if this is correct or not?
Thanks.
 
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If you assume the acceleration is constant then your answer is incorrect. Perhaps if you show your working we could correct you. Also note that the acceleration should be negative since the puck's velocity is decreasing.
 
i used the formula vf-vi/tf-ti to get that answer
 
petal5 said:
i used the formula vf-vi/tf-ti to get that answer
And how did you calculate the time?
 
i divided 15m by 5.5m.however,i now realize this is wrong as the 5.5m. is INITIAL speed
 
petal5 said:
i divided 15m by 5.5m.however,i now realize this is wrong as the 5.5m. is INITIAL speed
Indeed, so if we assume the acceleration to be uniform how does the displacement, initial and final velocity relate to the acceleration?
 
sorry,i've no idea.
 
would it be this equation:v^2=vo^2+2a(x-xo)
 
  • #10
petal5 said:
would it be this equation:v^2=vo^2+2a(x-xo)
Indeed it would, all that is left to do is some algebraic manipulation and to substitute in your values.
 
  • #11
thanks for all your help.i'm now getting -1.01m/s^2
 
  • #12
petal5 said:
thanks for all your help.i'm now getting -1.01m/s^2
Looks about right to me. My pleasure.
 
  • #13
Thanks so much,you're a great teacher!
 
  • #14
petal5 said:
Thanks so much,you're a great teacher!
Thank you very much :smile: I do my best.
 

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