Sled Riding

A sled and its rider are moving at a speed of 4.0 m/s along a horizontal stretch of snow, as Figure 4.24a illustrates. The snow exerts a kinetic frictional force on the runners of the sled, so the sled slows down and eventually comes to a stop. The coefficient of kinetic friction is 0.050. What is the displacement x of the sled?

Figure 4.24

(a)

The moving sled decelerates because of the kinetic frictional force.

(b)

Three forces act on the moving sled, the weight of the sled and its rider, the normal force , and the kinetic frictional force . The free-body diagram for the sled shows these forces.

Reasoning As the sled slows down, its velocity is decreasing. As our discussions in Chapters 2 and 3 indicate, the changing velocity is described by an acceleration (which in this case is a deceleration since the sled is slowing down). Assuming that the acceleration is constant, we can use one of the equations of kinematics from Chapter 3 to relate the displacement to the initial and final velocities and to the acceleration. The acceleration of the sled is not given directly. However, we can determine it by using Newton's second law of motion, which relates the acceleration to the net force (which is the kinetic frictional force in this case) acting on the sled and its mass.
Knowns and Unknowns The data for this problem are listed in the table:

Description	Symbol	Value	Comment
Explicit Data
Initial velocity	v_0x	+4.0 m/s	Positive, because the velocity points in the +x direction. See drawing.
Coefficient of kinetic friction	μ_k	0.050
Implicit Data
Final velocity	v_x	0 m/s	The sled comes to a stop.
Unknown Variable
Displacement	x	?

Modeling the Problem

Displacement To obtain the displacement x of the sled we will use Equation 3.6a from the equations of kinematics:

Solving for the displacement x gives the result shown at the right. This equation is useful because two of the variables, v_0x and v_x, are known and the acceleration a_x can be found by applying Newton's second law to the accelerating sled (see Step 2).

Newton's Second Law Newton's second law, as given in Equation 4.2a, states that the acceleration a_x is equal to the net force ΣF_x divided by the mass m:

The free-body diagram in Figure 4.24b shows that the only force acting on the sled in the horizontal or x direction is the kinetic frictional force

. We can write this force as -f_k, where f_k is the magnitude of the force and the minus sign indicates that it points in the -x direction. Since the net force is ΣF_x = -f_k, Equation 4.2a becomes

This result can now be substituted into Equation 1, as shown at the right.

Kinetic Frictional Force We do not know the magnitude f_k of the kinetic frictional force, but we do know the coefficient of kinetic friction μ_k. According to Equation 4.8, the two are related by


	(4.8)

where F_N is the magnitude of the normal force. This relation can be substituted into Equation 2, as shown at the right. An expression for F_N will be obtained in the next step.

Normal Force The magnitude F_N of the normal force can be found by noting that the sled does not accelerate in the vertical or y direction (a_y = 0 m/s²). Thus, Newton's second law, as given in Equation 4.2b becomes

There are two forces acting on the sled in the y direction, the normal force

and its weight

[see part (b) of the drawing]. Therefore, the net force in the y direction is

where W = mg (Equation 4.5). Thus, Newton's second law becomes

This result for F_N can be substituted into Equation 4.8, as shown at the right.

Solution Algebraically combining the results of each step, we find that

Note that the mass m of the sled and rider is algebraically eliminated from the final result. Thus, the displacement of the sled is