A 52.0 kg ice skater is gliding along the ice, heading due north at 4.50 m/s. Th

Question

A 52.0 kg ice skater is gliding along the ice, heading due north at 4.50 m/s. The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but ?k-0. Suddenly, a wind from the northeast exerts a force of 4.20 N on the skater at Use work and energy to find the skater's speed after gliding 100 m in this wind. Express your answer with the appropriate units. -296.98 J Submit Previous Answers Request Answer x Incorrect: Thy Again; 29 attempts remaining Enter your answer using dimensions of velocity. What is the minimum value of us 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 = 4.20 cos 45 = 2.96 N
W = F * d = 2.96 N * 100 m  
W = 296 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 [(4.5 m/s)^2 - (2*296 N*m/52.0 kg)]
v2 = 3449 m/s

PartB: 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 = (4.20 N * cos 45 )/ (52.0 kg * 9.81 m/s^2)
Us = 0.00582

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