A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . T

Question

A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but Uk = 0. Suddenly, a wind from the northeast exerts a force of 3.70 N on the skater. a.) Use work and energy to find the skater's speed after gliding 100 m in this wind. b.) What is the minimum value of Ug that allows her to continue moving straight north? A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but Uk = 0. Suddenly, a wind from the northeast exerts a force of 3.70 N on the skater. a.) Use work and energy to find the skater's speed after gliding 100 m in this wind. b.) What is the minimum value of Ug that allows her to continue moving straight north?

Explanation / Answer

Equations:
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 = 3.70 cos 45 = 2.62 N
W = F * d = 2.62 N * 100 m
W = 261.6 N*m

KE1 = Initial kinetic energy
KE2 = kinetic energy after 100 m

The energy after 100 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 [(4.1 m/s)^2 - (2*261.6 N*m/54.0 kg)]
v2 = 2.668 m/s

Question 2: Min Us
Ff = force of friction
Us = coefficient of static friction
N = Normal force = weight of skater

Ff = Us*N
solve for Us
Us = Ff/N
Us = (3.70 N * cos 45 )/ (54.0 kg * 9.81 m/s^2)
Us = 0.00494

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