Maths question

[Jandywa]

A hockey puck starts sliding across some horizontal rough ice with an initial speed
of 9 m/s. It slides for 45m before stopping. Find the coefficient of friction

[2 sMoKiN bArReLs]

45 miles is an awful long rink?

[ATTAK Z]

I bet it's 5 but can't prove it

[Simpo Two]

And 9 miles per second is awfully fast

Damn Frenchies.

I have no idea, 45/9?

[Jandywa]

2 sMoKiN bArReLs said:

45 miles is an awful long rink?

i believe it is regulation size

[ATTAK Z]

Jandywa said:

......... Find the coefficient of friction

I'd be asking which stupid bar-steward lost it in the first place

[SMGB]

0.1

[Jandywa]

SMGB said:

0.1

i think you might be right

[2 sMoKiN bArReLs]

But science friction burns your fingers

(You need to be a certain age to get that )

[fido]

I get ..

0.9 / g which is close to 0.1

[tank slapper]

fido said:

I get ..

0.9 / g which is close to 0.1

I agree.

[Simpo Two]

Jandywa said:

2 sMoKiN bArReLs said:

45 miles is an awful long rink?

i believe it is regulation size

Wait a minute, if you mean that's the size of the ice rink then it might have gone ker-thunk into the side, in which case it still had speed after 45m so the coefficient of slideyness is lower than you thought. It can't travel more than 45m no matter how hard it is hit (until it punches through the side of course)

[paranoid airbag]

http://en.wikipedia.org/wiki/Formulas_for_constant... is your friend, specifically the second equation, often written as:

v^2 = u^2 + 2*a*s

since by stating 'find the coefficient of friction' the question is sort of implying you can assume constant deceleration. v is your starting velocity (9), u the final velocity (0), a is acceleration (what you want to find) and s the distance. Rearrange and you should get a = 0.9, you know from newton's second law that

Force (F) = mass (m) * a

and the coefficient of friction (mu) is the ratio of the friction forcce (m*a, above) to the contact force (equal here to the weight, which is equal to m*g):

mu = m*0.9/m*g = 0.9/g

[tank slapper]

The alternate method to that is to use the fact that all the kinetic energy has been dissipated by the time the puck comes to rest:

Energy is force acting over a distance, so E = F*d.

Kinetic energy is E = 1/2*m*v^2

so 1/2*m*v^2 = F*d

The frictional force is F = μ*N where N = m*g so we get

1/2*m*g = μ*m*g*d

m cancels out

1/2*g = μ*g*d

so μ = v^2 / 2*g*d

substitute the known values and you get μ=0.9/g.

[TwigtheWonderkid]

2 sMoKiN bArReLs said:

But science friction burns your fingers

(You need to be a certain age to get that )

A certain age, or from Swindon.