Maths question
Discussion
Jandywa said:
2 sMoKiN bArReLs said:
45 miles is an awful long rink?
i believe it is regulation size 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
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
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.
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.
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