Work done by friction on a hockey puck

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AI Thread Summary
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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TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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