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?

Answer 1

Answer:

a. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

$W = F\times d$

$KE = 0.5\times m\times v^2$

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

$W = F \times d = 2.62 N \times 100 m$

$W = 261.6 N\times m$

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

$0.5 m\times v2^2 = 0.5 m\ v1^2 - W$

Now solve for v2

$v2 = \sqrt{v1^2 - {\frac{2W}{M}}}$

$= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}$

= 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,

$Ff = Us\times N$

Now solve for Us

$= \frac{Ff}{N}$

$= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}$

= 0.00494