Work done by friction on a hockey puck

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    Friction Work
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Homework Help Overview

The discussion revolves around calculating the work done by friction on a hockey puck moving across a rough surface, where the coefficient of kinetic friction varies with position. The original poster questions the accuracy of their calculation compared to an answer key.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of work done by friction, referencing the relevant equations and questioning the values provided in the answer key. There is a mention of using Wolfram Alpha for verification of calculations.

Discussion Status

The conversation is ongoing, with participants expressing frustration over discrepancies between their calculations and the answer key. Some participants are exploring the implications of the mass of the hockey puck and its relevance to the problem.

Contextual Notes

Participants note a potential typo regarding the mass of the hockey puck and express uncertainty about the typical weight of hockey pucks, indicating a lack of familiarity with the sport.

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Homework Statement
A hockey puck of mass 0.17 kg is shot across a rough floor with the roughness different at different places, which can be described by a position-dependent coefficient of kinetic friction. For a puck moving along the x-axis, the coefficient of kinetic friction is the following function of x, where x is in m: u(x) = 0.1 + 0.05x. Find the work done by the kinetic frictional force on the hockey puck when it has moved from x = 0 to x = 2
Relevant Equations
[math]W = \int_0^2 F(x) \, dx[/math]
The answer key claims the answer is -0.9 J.

By my math it should be -0.501 J, is that incorrect?
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I_Try_Math said:
Homework Statement: A hockey puck of mass 0.17 kg is shot across a rough floor with the roughness different at different places, which can be described by a position-dependent coefficient of kinetic friction. For a puck moving along the x-axis, the coefficient of kinetic friction is the following function of x, where x is in m: u(x) = 0.1 + 0.05x. Find the work done by the kinetic frictional force on the hockey puck when it has moved from x = 0 to x = 2
Relevant Equations: [math]W = \int_0^2 F(x) \, dx[/math]

The answer key claims the answer is -0.9 J.

By my math it should be -0.501 J, is that incorrect?
View attachment 339633
You mean " by Wolfram Alpha's math" its ##-0.501 \rm{J}##. Wolframs math looks correct to me also...
 
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erobz said:
You mean " by Wolfram Alpha's math" its ##-0.501 \rm{J}##. Wolframs math looks correct to me also...
Thanks, it's frustrating trying to learn when you can't check your work against the answer key.
 
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haruspex said:
I assume the "0.17kg" was a typo.
0.17x9.8=1.67!?!!

But yeah I got no clue how heavy is a hockey ball, hockey isn't popular here in Greece.
 
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Delta2 said:
0.17x9.8=1.67!?!!

But yeah I got no clue how heavy is a hockey ball, hockey isn't popular here in Greece.
I had to look it up. Google "how heavy is a hockey puck".
https://en.wikipedia.org/wiki/Hockey_puck said:
Ice hockey requires a hard disk of vulcanized rubber. A standard ice hockey puck is black, 1 inch (25 mm) thick, 3 inches (76 mm) in diameter, and weighs between 5.5 and 6 ounces (156 and 170 g);[6] some pucks are heavier or lighter than standard (see below). Pucks are often marked with silkscreened team or league logos on one or both faces.[6] Pucks are frozen before the game to reduce bouncing during play.[6]
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