Mechanics question-Momentum&F=ur

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The discussion revolves around solving a mechanics problem involving an ice hockey puck's motion. The puck, with a mass of 1.2 kg, must accelerate uniformly to cover 25 m in 8 seconds, requiring an acceleration of -0.78125 m/s² and a minimum force of 0.9375 N applied by the player. The momentum at the end of the 8 seconds is calculated to be 7.5 kg·m/s. Additionally, the puck travels another 8 m before coming to rest, leading to a calculated frictional force of -0.7102 N and a coefficient of kinetic friction of approximately -0.0603. The conversation highlights confusion regarding the negative acceleration and the transition from no friction to the effects of friction in the problem.
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[SOLVED] Mechanics question-Momentum&F=ur

Homework Statement



An ice hockey player is required to strike a puck having a mass of 1.2 kg, so that it is
accelerated uniformly from rest across a smooth ice rink surface so as to cover a distance of 25 m in 8 s.
The puck travels another 8 m before coming to a rest. Determine (a) the
acceleration that the puck has to experience, (b) the minimum force the ice hockey player has to apply to the puck to achieve this acceleration, (c) the momentum of the puck at the end of the 8 s period, (d) the frictional force that brings the puck to rest, and (e) the coefficient of kinetic friction between the ice and the puck.

Homework Equations



s=ut+1/2at^2
F=\mu R
f=ma

The Attempt at a Solution



(a)use the formula shown above u=0 t=8s s=25m therefore simply substitue into equation a=-0.78125
(b)use 3rd equation m=1.2kg a=-0.78125 f=0.9375N
(c)momentum is g=mv . Once again substitute values into equation. m=1.2kg v=6.25m/s found out by using s=1/2 (u+v)t g=7.5kg m/s
(d)&(e). Both use the 2nd formula but I'm unsure as how to go about it. I know R=mg. but F can't be the same as in f=ma can it?

Also check over the other stuff too please for any mistakes thanks.
 
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For (d) you know the puck comes to rest in 8m from the speed you calculated above. You need to use the kinematic equations to find a and then F. Once you find F you can use the second equation to get \mu.
 
oh ok. that means that the new acceleration a_2=-0.59185606 found by using the equation v^2=u^2 + 2as (u=6.25m/s s=33 v=0) when inserted into f=ma gives me f_2=-0.710227272 then inserting that into F=\mu R gives me \mu=-0.060331912.

Is the working out for the other parts of the question right too?
 
Last edited:
Seems to be but I'm not sure why your acceleration in the first is negative. The question is a bit weird looking at it again. It assumes no friction for the first part then suddenly friction comes into slow it down. Oh well. :smile:
 
i probably messed up somewhere. Thanks for the help much appreciated.
 
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