A skier slides horizontally along the snow for a distance of 21 meters before coming to rest The coefficient of kinetic friction is 0.050 how fast was the skier going?

Answer 1

For the skier to come to rest, it has to lose all its kinetic energy. Ignoring the drag force caused by air, kinetic energy is strictly lost through friction with the snow. The kinetic friction force Fk is mu, the kinetic friction coefficient, times the normal to the surface, in that case the total weight M*g of the skier. The total work the friction does against the skier is thus Fk times x, where x is the distance over which the friction force acted. Remembering the formula for kinetic energy, the principle of energy conservation thus dictates the following equality :

1/2*M*v^2 = mu*M*g*x

We see the total mass of the skier becomes irrelevent since it cancels out of the right and left hand side of the equality. Therefore, the speed at which the skier was going before starting to slow down is

v = sqrt(2*mu*g*x)

and since

mu = 0.050

g = 9.81 m/s^2

x = 21 m

then v = 4.539 m/s, which is equivalent to 16.332 km/h once you multiply by 3.6 km/h per m/s.