Question

A 48 kg ice skater is gliding along the ice, heading due north at 5.0 m/s. The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but

Answer 1

W = F*d
KE = 0.5*m*v^2

Since the wind fails to move the skater any to the east there is no work done in this direction. All the work is done in the N-S direction. So find the component of the Force in that direction.

F = 4 cos 45 = 2.76 N
W = F * d = 2.76 N * 90 m
W = 248 N*m

KE1 = Initial kinetic energy
KE2 = kinetic energy after 90 m

The energy after 90 meter equals the initial kinetic energy minus the energy lost to the work performed by the wind on the skater.

KE2 = KE1 - W
0.5 m*v2^2 = 0.5 mv1^2 - W
solve for v2
v2 = sqrt[ v1^2 - 2W/m]
v2 = sqrt [(5 m/s)^2 - (2*248 N*m/48 kg)]
v2 = 3.83 m/s