Need to calculate the friction coefficient

[MisterP]

Homework Statement

Hello. I have slope = 45 degrees, at distance 36.4cm (0.364m) object gains 2m/s speed. Need to calculate friction coefficient. Correct answer is ( 0,2). How to calculate?

Homework Equations

1) t = s/v "t" - time; "s" - distance "v" - speed
2) S = (a*t²)/2 "S" - distance; "a" - acceleration; "t" - time (found this formula)
3) a = g(sin "angle" - mi*cos "angle") "a"-acceleration; "g" - gravity constant (9.81), "angle" = 45 degrees; mi - coefficient I need to find!

The Attempt at a Solution

So I used formulas: 1) to calculate time t = s/v = 0.364/2 = 0.182s
2) then to calculate acceleration.. S = (a*t²)/2 =
0,364 = (a*0.182²)/2 = 5.15m/s²
3)then to calculate "mi" ; 5.15 = 9.81(sin "45^o" - mi*cos "45^o") =
5.15 = 6.92 - mi*6.92; mi = 0.25

Anything right here? :D English is not my native so, please, try to express yourself as easy as You can, thank You :)

 

[TSny]

Hello and welcome to PF!

Please give a complete statement of the problem. For example, it is not clear if the object started at rest.

The equation t = s/v is valid only if v is the average velocity that the object had while moving the distance s in time t.
The 2 m/s is the instantaneous velocity of the object at the instant it has traveled the distance of 36.4 cm

Otherwise, I think your work looks good.

For constant acceleration, there is a nice formula that relates the final velocity to the initial velocity, the acceleration, and the distance traveled. If you have covered that formula, then you can use it to find the acceleration without needing to find the time t.

0,364 = (a*0.1822)/2 = 5.15m/s2

Certainly, 0.364 does not equal 5.15 m/s². The second equals sign is not valid. I think you meant to say that you solved the first equality for the acceleration and that you got a = 5.15m/s².

 

[neilparker62]

Work done by non conservative forces = change in mechanical energy.

 

[MisterP]

I will write down that example and try my best to translate it:
Object is sliding down the slope, that is 45 degrees (to horizon). By sliding down 36.4 cm, object reaches speed - 2m/s. Calculate friction coefficient. Correct answer is 0.2

 

[TSny]

I will write down that example and try my best to translate it:
Object is sliding down the slope, that is 45 degrees (to horizon). By sliding down 36.4 cm, object reaches speed - 2m/s. Calculate friction coefficient. Correct answer is 0.2

OK. I think you will get the correct answer if you find the correct time t. If the object starts at rest and has a final speed of 2 m/s, what is its average speed? Then, t can be found from distance = (average speed) x (time).

@neilparker62 has pointed out an entirely different approach to the problem which is nice, if you are familiar with energy and work concepts.

 

[haruspex]

If the object starts at rest and has a final speed of 2 m/s, what is its average speed?

Unfortunately that approach requires one to assume constant acceleration, which is not given.
The work/energy method avoids that.

 

[TSny]

Unfortunately that approach requires one to assume constant acceleration, which is not given.
The work/energy method avoids that.

True. Although it is not given that the acceleration is constant, the OP had already stated the correct formula for the acceleration. See relevant equation #3. So I assumed that the OP realized the acceleration is constant.

 

[MisterP]

What if it is not meant to be as constant acceleration, any choices to solve that problem?

 

[neilparker62]

See post #3. In this case the "non conservative force" is friction.

 

[neilparker62]

Taking OP's equation 3 and replacing the canceled 'm' (mass of object):

$$ ma = mg(sinθ-μcosθ) $$
Replace a with $\frac{{v_f}^2}{2Δx}$ as suggested by TSny in post #2
$$ m{v_f}^2/(2Δx) = mgsinθ-μmgcosθ $$
$$ ½m{v_f}^2 = mgsinθΔx-μmgcosθΔx $$
Re-arrange: $$ -μmgcosθΔx = ½m{v_f}^2 - mgΔxsinθ $$
Hey presto: work done by non conservative forces = change in mechanical energy !

 

[MisterP]

What is "Δx" and "ƒ" ?

 

[neilparker62]

What is "Δx" and "ƒ" ?

Δx is the displacement (36,4 cm) and v_f is 'final velocity' which is 2m/s in this problem.

 

[Orion73]

Assuming that the object started at rest, calculate the net acceleration using v^2=u^2+2as.Now divide g into horizontal and vertical components. Take frictional force along the plane opposite to the direction of motion. Net acceleration a=g.cos 45°-f, where first is frictional retardation. Find f. To find coefficient of friction, divide f by g. sin 45°...obtained answer is 0.2

 

[MisterP]

Taking OP's equation 3 and replacing the canceled 'm' (mass of object):

$$ ma = mg(sinθ-μcosθ) $$
Replace a with $\frac{{v_f}^2}{2Δx}$ as suggested by TSny in post #2
$$ m{v_f}^2/(2Δx) = mgsinθ-μmgcosθ $$
$$ ½m{v_f}^2 = mgsinθΔx-μmgcosθΔx $$
Re-arrange: $$ -μmgcosθΔx = ½m{v_f}^2 - mgΔxsinθ $$
Hey presto: work done by non conservative forces = change in mechanical energy !

I tried to calculate and cannot get to coefficient: 0.2
-μ * 9.81 * √2/2* 0.364 = 1/2*2 - 9.81*0.364 * √2/2
I get μ = 0.6

 

[MisterP]

Assuming that the object started at rest, calculate the net acceleration using v^2=u^2+2as.Now divide g into horizontal and vertical components. Take frictional force along the plane opposite to the direction of motion. Net acceleration a=g.cos 45°-f, where first is frictional retardation. Find f. To find coefficient of friction, divide f by g. sin 45°...obtained answer is 0.2

What do you mean : divide g into horizontal and vertical components?

 

[neilparker62]

I tried to calculate and cannot get to coefficient: 0.2
-μ * 9.81 * √2/2* 0.364 = 1/2*2 - 9.81*0.364 * √2/2
I get μ = 0.6

-μ * 9.81 * √2/2* 0.364 = 1/2*2^2 - 9.81*0.364 * √2/2

 

[MisterP]

-μ * 9.81 * √2/2* 0.364 = 1/2*2^2 - 9.81*0.364 * √2/2

Yes, it works now, thank You :)

 

[MisterP]

Could You tell me if there is any compilation table or something where I could find all formulas on basic physics ( mechanical, molecular, electrical etc)

 

[neilparker62]

Here schools use a data sheet such as the following but I don't know how relevant it will be elsewhere:

http://www.past-exam-papers.co.za/past-exam-paper-downloads/Exam-E-Chemistry-IEB-Datasheet.pdf

 

[sojsail]

Could You tell me if there is any compilation table or something where I could find all formulas on basic physics ( mechanical, molecular, electrical etc)

I hope this table (and the pages that the links take you to) will help: http://www.softschools.com/formulas/physics/friction_formula/32/

 

[MisterP]

Thank You, yes. :)