Work done by friction on a hockey puck

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The discussion centers around the calculation of work done by friction on a hockey puck, with the answer key stating -0.9 J, while one participant calculates it as -0.501 J using the coefficient of kinetic friction function. There is agreement that the calculation from Wolfram Alpha, which also yields -0.501 J, appears correct. Frustration is expressed over discrepancies between personal calculations and the answer key, highlighting the challenge of verifying work. Additionally, there is a brief exchange about the weight of a hockey puck, clarifying that it typically weighs between 156 and 170 grams. Accurate calculations and understanding of physics principles are essential for resolving such discrepancies.
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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.
 
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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Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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