If an ice skater moving at 10 meters coasts to a stop in 100 meters on a smooth ice surface then what is the coefficient between the ice and the skates?

Answer 1

The easiest way to solve this problem is to use the conservation of energy, assuming there are no external forces (such as air drag) acting on the skater/ice system.

The two forms of energy relevant to this problem are kinetic energy K and thermal energy Et.

Since energy is conserved, the energy of the system must be the same before and after the skater came to a stop.

Ki + Eti = Kf + Etf

where K = 1/2mv2 and Et = fd

Thermal energy Et is equal to the product of the frictional force and the distance upon which it acts, where the frictional force f can be rewritten as a product of the coefficient of kinetic friction µ and the normal force n.

f = µn

Since there are no other forces acting on the skater in the vertical direction, the normal force n must be equal to the gravitational force Fg, which is the product of the mass of the skater times the acceleration due to gravity.

Fg = mg

So we get that the thermal energy can be rewritten as

Et = µmgd.

Substituting all of this back into the original equation yields:

1/2mv2i + µmgdi = 1/2mv2f + µmgdf

Notice that each term contains mass m, so they cancel out. You are left with

1/2v2i + µgdi = 1/2v2f + µgdf

The thermal energy at the very beginning of the problem is zero, and the kinetic energy at the very end is zero (the skater has stopped), so those terms also drop out.

1/2v2i + 0 = 0 + µgdf

1/2v2i = µgdf

Solving for µ yields:

µ = (v2i)/(2gd)

Knowing the acceleration due to gravity to be 9.8 m/s2 and having been given the initial velocity vi = 10 m/s and the distance d = 100 meters, one can simply plug in the values.

µ = 102/2*9.8*100
µ = 0.05

(The coefficient of friction has no units)