A $50kg$ice skater is gliding along the ice, heading due north at $4.0m/s$. The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but ${\mu }_k=0$. Suddenly, a wind from the northeast exerts a force of $4.0N$ on the skater.

a. Use work and energy to find the skater’s speed after gliding $100m$ in this wind.

b. What is the minimum value of ${\mu }_s$ that allows her to continue moving straight north?

Work done is equal to the product of force and displacement in the direction of force and is represented as,

$W=Fd........................\left(1\right)$

according to work energy theorem, the change in kinetic energy is equal to work done. it is shown as,

role="math" localid="1647790208972" $$\begin{gathered} W_{net}=K{E}_2-K{E}_1 \\ =\frac{1}{2}m({v_2}^2-{v_1}^2).............................\left(2\right) \end{gathered}$$

The coefficient of friction is the ratio between the frictional force and normal reaction force. it is expressed as,

$\mu \frac{f}{N}..................\left(3\right)$

The weight of body is represent as

$w=mg$

Consider the force acting along north and east as positive and the force acting along south and west is negative.

Draw the free body diagram

Here , F is the force applied by the wind on the skater, f is the force by kinetic friction and f_s is static friction force.

The component of force acting southwards is

$F_{south}=F\cos \left(45°\right)$

Substitute

role="math" localid="1647790667372" $$\begin{gathered} 4NforF \\ {F}_{south}=4N\cos \left(45°\right) \end{gathered}$$

Calculate the work done by wind,

$$\begin{gathered} W_{wind}=-(4.0N)\cos \left(45°\right)(100m) \\ =-282.84J \end{gathered}$$

The coefficient of kinetic friction is zero, and hence friction does no work in the motion of skater.

Apply the work- energy theorem, to calculate the velocity after gliding $100m$in the wind. Substitute the values

$$\begin{gathered} -282.84J=\frac{1}{2}m({v_2}^2-{(4.0m/s)}^2)v_2=2.164m/s \\ {v}_2=2.12m/s \end{gathered}$$

According to the free body diagram of the skater,

$f_s=F\sin \left(45°\right)$

Substitute $4.0NforF.$

$$\begin{gathered} f_s=(4N)\sin \left(45°\right) \\ {f}_s=2.83N \end{gathered}$$

the normal reaction from the ground is equal to the weight of the skater, as the direction is perpendicular to the Earth's surface. The expression for normal reaction is written as,

role="math" localid="1647791215973" $$\begin{gathered} N=mg \\ N=50kg(9.8m/s^2) \\ N=490N \end{gathered}$$

The coefficient of static friction is

$$\begin{gathered} {\mu }_s=\frac{f_s}{N} \\ {\mu }_s=\frac{2.83N}{490N} \\ {\mu }_s=0.0058 \end{gathered}$$