91Ó°ÊÓ

The concept of frictional force is crucial in many physics problems. Frictional force resists the motion of an object and acts in the opposite direction.

When you push a book across a table, it eventually stops because of friction. In the given exercise, the ice skater slows down and stops due to the slight frictional force between the ice and the skates.

Frictional force can be calculated using the formula:

\[F_f = \, \mu\,^{\text{kin}} \times F_N\]
The coefficient of kinetic friction (\( \mu\,^{\text{kin}} \)) measures how slippery the surface is. A higher value means more friction.

• For ice, \( \mu\,^{\text{kin}} \)is usually low because ice is slippery.


• The normal force (\( F_N \)) is the force perpendicular to the surface. For a skater on a flat surface, it equals:\( F_N = mg \)where \( m \) is the mass and \( g \) is gravitational acceleration (\( 9.8 \, \text{m/s}^2 \)).

When you multiply the coefficient of friction by the normal force, you get the frictional force slowing down the skater.
This force determines how quickly she will come to rest.

Kinetic energy (\( K \)) is the energy of motion. Anything moving has kinetic energy. The skater in the exercise initially has kinetic energy because she's moving at 2 m/s:

\[ K_i = \frac{1}{2}mv_1^2 \]where

• \( m \) is the mass of the skater.


• \( v_1 \) is her speed (2 m/s).

Her kinetic energy changes because of work done by the friction. According to the work-energy theorem, the work done (\( W \)) on an object is equal to its change in kinetic energy. For the skater:

\[ W = \Delta K = K_f - K_i \]Since the skater comes to rest, her final kinetic energy (\( K_f \)) is 0. Thus, the net work done is:

\[ W = 0 - \frac{1}{2}mv_1^2 \]
This work is done by the frictional force acting over some distance (\( d \)):

\[ W = F_f \times d \]
By equating the two expressions for work, you can solve for the distance (\( d \)) the skater travels before stopping:

\[ -\frac{1}{2}mv_1^2 = \, \mu\,^{\text{kin}} mg d \]
Simplifying, we obtain:

\[ d = \frac{v_1^2}{2 \, \mu\,^{\text{kin}} g} \]
Insert the values to find that the skater travels approximately 1.70 meters before stopping.

Free-body diagrams are visual tools to depict the forces acting on an object. They simplify complex problems by breaking down the forces involved.

In the skater example, we need to show the horizontal and vertical forces acting on her:


• Draw a dot or box to represent the skater.


• Draw arrows originating from the dot to represent forces.


For the vertical forces:
• The gravitational force (\( F_g \)) acts downwards.


• The normal force (\( F_N \)) acts upwards and is equal in magnitude to the gravitational force on a flat surface:
Simplified, \( F_N = mg \)where \( m \) is mass and \( g \) is gravitational acceleration.For the horizontal force:
• The frictional force (\( F_f \)) acts opposite to the direction of motion and slows down the skater.
In the diagram:
• \( F_g \) is shown as an arrow pointing down.


• \( F_N \) is shown as an arrow pointing up. They should be equal in length as \( F_N = F_g \)


• \( F_f \) is shown as an arrow pointing opposite to the direction of motion.
This simple diagram helps to visualize and solve the problem by mapping out the forces clearly.