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I'chiliad aware that this is extremely basic just I've forgotten how to go the coefficient of friction when given a forward strength and acceleration.

Can someone depict to me the algorithm for solving the following case question?

Mass = 50kg

Forward Force Acting on Mass = 100 N

Acceleration of Mass = 0.1m/s^2

Coefficient of friction?

EDIT: Am I right in thinking information technology is 0.2?

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Qmechanic♦

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asked Jun 10, 2015 at 13:34

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  • $\begingroup$ Have you revisted an introductory physics textbook, or looked online at recourses like hyperphysics? $\endgroup$

    Jun x, 2015 at 13:40

  • $\begingroup$ I take but I didn't see annihilation that regarded getting the coefficient of friction when acceleration is included. I am having somewhat of a mental-bare at the moment and just desire to make sure I know the steps needed every bit I accept a physics test later today. EDIT: My mental fog is clearing and I am outset to realize ridiculously easy this is. I'd still like an answer to assure I am correct but I'k adequately sure I can figure this out on my own $\endgroup$

    Jun x, 2015 at thirteen:42

  • $\begingroup$ At that place are two forces, the forward strength and friction. The departure of these forces is the cyberspace force equal to the product of mass and dispatch. From this you get the friction strength. From this split up then normal force, mg to become them friction force. $\endgroup$

    Jun 10, 2015 at 13:53

  • $\begingroup$ It is 0.ii right? $\endgroup$

    Jun ten, 2015 at 13:54

  • $\begingroup$ Really 0.1938...., only rounding to ii decimals or using m=10, you get 0.19 $\endgroup$

    Jun 10, 2015 at xiii:59

2 Answers 2

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Do a free body diagram and you will find for a horizontal plane that

$$ F - \mu m g = m a $$

$$ (100) - \mu (l) (nine.80665) = (50) (0.1) $$

$$\boxed{ \mu = \frac{(100)-(0.ane)(50)}{(l)(9.80665)} = 0.1937\ldots }$$

answered Jun ten, 2015 at 16:37

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  • $\begingroup$ Ah yes. Immediately substitute values into the starting equation and avoid the algebra necessary to dissever the unknown variable. THAT technique is why most U.S. high school students are practically algebraically illiterate. $\endgroup$

    Feb 24, 2019 at 21:26

  • $\begingroup$ @DavidWhite - I really I didn't, since the final expression was already solved for $\mu$ and the commutation was the final step. The deviation is if I carry the symbols vs. their value around. $\endgroup$

    Feb 25, 2019 at 0:05

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Newton's second constabulary in the 10-management. At that place are only 2 forces that together equal the $ma$, and since you know ane of them, it is directly forward from here.

answered Jun x, 2015 at 14:41

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