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. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.

F = 3.70 cos 45 = 2.62 N

We know that

KE1 = Initial kinetic energy

KE2 = kinetic energy following 100 m

The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.

So, the equation is

KE2 = KE1 - W

Now solve for v2

= 2.668 m/s

b. Now the minimum value of Ug is

As we know that

Ff = force of friction

Us = coefficient of static friction

N = Normal force = weight of skater

So,

Now solve for Us

= 0.00494

answer: the direction of the wind is expressed as the direction from which the wind is blowing. for example, easterly winds blow from east to west, while westerly winds blow from west to east.

explanation:

i believe the answer is a hope it and is correct