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A sled and its rider are moving at a speed of  along a horizontal stretch of snow, as Figure 4.24 a 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?
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 to the mass of the sled (plus rider).
Knowns and Unknowns The data for this problem are listed in the table:
Description |
Symbol |
Value |
Comment |
Explicit Data |
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Initial velocity |
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Positive, because the velocity points in the  direction. See the drawing. |
Coefficient of kinetic friction |
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0.050 |
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Implicit Data |
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Final velocity |
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The sled comes to a stop. |
Unknown Variable |
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Displacement |
x |
? |
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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,  and  , are known and the acceleration  can be found by applying Newton's second law to the accelerating sled (see Step 2). |
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 Newton's Second Law Newton's second law, as given in Equation 4.2a, states that the acceleration is equal to the net force  divided by the mass m:
The free-body diagram in Figure 4.24 b 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  , where  is the magnitude of the force and the minus sign indicates that it points in the  direction. Since the net force is  , Equation 4.2a becomes
This result can now be substituted into Equation 1, as shown at the right. |
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 Kinetic Frictional Force We do not know the magnitude of the kinetic frictional force, but we do know the coefficient of kinetic friction  . According to Equation 4.8, the two are related by
where  is the magnitude of the normal force. This relation can be substituted into Equation 2, as shown at the right. An expression for will be obtained in the next step. |
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 Normal Force The magnitude of the normal force can be found by noting that the sled does not accelerate in the vertical or y direction  . 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 Figure 4.24 b). Therefore, the net force in the y direction is
where  (Equation 4.5). Thus, Newton's second law becomes
This result for can be substituted into Equation 4.8, as shown at the right. |
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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
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