A skater of mass 60 kg has an initial velocity of 12 m/s. He slides on ice where the frictional force is 36 N. How far will the skater slide before he stops?

d = 120 [m]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. Where the energy in the final state (when the skater stops) is equal to the sum of the mechanical energy in the initial state plus the work done on the skater in the initial state.

The mechanical energy is equal to the sum of the potential energy plus the kinetic energy.  As the track is horizontal there is no unevenness, in this way, there is no potential energy.

E₁ + W₁₋₂ = E₂

where:

E₁  = mechanical energy in the initial state [J] (units of Joules)

W₁₋₂ = work done between the states 1 and 2 [J]

E₂  = mechanical energy in the final state = 0

E₁ = Ek = kinetic energy [J]

E₁ = 0.5*m*v²

where:

m = mass = 60 [kg]

v = initial velocity = 12 [m/s]

Now, the work done is given by the product of the friction force by the distance. In this case, the work is negative because the friction force is acting in opposite direction to the movement of the skater.

W₁₋₂ = -f*d

where:

f = friction force = 36 [N]

d = distance [m]

Now we have:

0.5*m*v² - (f*d) = 0

0.5*60*(12)² - (36*d) = 0

4320 = 36*d

d = 120 [m]