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

terd muffin is the correct answer

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000