Problem 22 - University Physics with Modern Physics

Question

Use the work\(-\)energy theorem to solve each of these problems. You can use Newton's laws to check your answers. (a) A skier moving at 5.00 m/s encounters a long, rough horizontal patch of snow having a coefficient of kinetic friction of 0.220 with her skis. How far does she travel on this patch before stopping? (b) Suppose the rough patch in part (a) was only 2.90 m long. How fast would the skier be moving when she reached the end of the patch? (c) At the base of a frictionless icy hill that rises at 25.0\(^\circ\) above the horizontal, a toboggan has a speed of 12.0 m/s toward the hill. How high vertically above the base will it go before stopping?

Step-by-Step Solution

Answer

(a) 5.76 m; (b) 3.92 m/s; (c) 7.34 m.

Step 1: Understanding the Work-Energy Principle

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. Mathematically, this can be expressed as \( W = rac{1}{2}mv^2_f - rac{1}{2}mv^2_i \), where \( W \) is the work done, \( v_f \) is the final velocity, \( v_i \) is the initial velocity, and \( m \) is the mass of the object.

Step 2: Part (a): Setting up the Equation for the Skier

For part (a), the skier moves with an initial speed \( v_i = 5.00 \, \text{m/s} \) and comes to a stop, so \( v_f = 0 \, \text{m/s} \). The work done is due to the kinetic friction force, which is \( W = -f_k \cdot d \), where \( f_k \) is the kinetic friction force and \( d \) is the distance traveled. The kinetic friction force is \( f_k = \mu_k mg \), where \( \mu_k = 0.220 \), and \( m \) is the skier's mass.

Step 3: Calculate the Skier's Stopping Distance

Using the work-energy principle, \( -f_k \cdot d = -\frac{1}{2}mv_i^2 \). Substituting \( f_k = \mu_k mg \), we get \( -\mu_k mgd = -\frac{1}{2}mv_i^2 \). Canceling \( m \) and solving for \( d \), we have \[ d = \frac{\frac{1}{2}v_i^2}{\mu_k g} = \frac{0.5 \cdot (5.00)^2}{0.220 \cdot 9.81} \approx 5.76 \, \text{m}. \]

Step 4: Part (b): Skier's Speed at the End of the Patch

For part (b), the initial speed \( v_i = 5.00 \, \text{m/s} \), and the patch length is \( 2.90 \, \text{m} \). Using \[ \frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 - f_k \cdot 2.90, \] we substitute \( f_k = \mu_k mg \), then \[ \frac{1}{2}v_f^2 = \frac{1}{2}v_i^2 - \mu_k g \cdot 2.90. \] Solving for \( v_f \), we have \[ v_f = \sqrt{v_i^2 - 2\mu_k g \cdot 2.90} = \sqrt{(5.00)^2 - 2 \cdot 0.220 \cdot 9.81 \cdot 2.90} \approx 3.92 \, \text{m/s}. \]

Step 5: Part (c): Setting up the Equation for the Toboggan

In part (c), the work done is against gravity as the toboggan moves up the icy hill. All kinetic energy is converted into potential energy, so \( \frac{1}{2}mv_i^2 = mgh \), where \( h \) is the height gain. With \( v_i = 12.0 \, \text{m/s} \), solving for \( h \) gives \[ h = \frac{v_i^2}{2g} = \frac{(12.0)^2}{2 \cdot 9.81} \approx 7.34 \, \text{m}. \]

Key Concepts

Kinetic Friction

Kinetic friction is the force that opposes the motion of two surfaces sliding past each other. When a skier moves across a rough patch of snow, kinetic friction plays a significant role in slowing them down. The force of kinetic friction (\( f_k \)) depends on two factors:

• the coefficient of kinetic friction (\( \mu_k \)): a measure of how "grippy" or "slippery" the surfaces are in contact. For example, a \( \mu_k \) of 0.220 indicates moderate slipperiness.
• the normal force (\( N \)): typically equivalent to the weight of the object on flat ground. It can be calculated using \( N = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity.

The kinetic friction force is given by \( f_k = \mu_k \cdot N \). This force is crucial in applying the work-energy theorem to determine how far an object, like the skier, travels before stopping.

Newton's Laws

Newton's Laws of Motion are fundamental principles that explain the relationship between the motion of an object and the forces acting on it. Here's how they relate to the scenario of a skier on snow:

• Newton's First Law: An object will remain at rest or in uniform motion unless acted upon by a net external force. For the skier, it's the force of kinetic friction that eventually brings her to rest.
• Newton's Second Law: This law states that the acceleration of an object is proportional to the net force acting on it and inversely proportional to its mass (\( F = ma \)). Here, \( f_k \), the force due to kinetic friction, causes a deceleration because it's acting opposite to the skier's motion.
• Newton's Third Law: For every action, there is an equal and opposite reaction. As the skier moves forward, the snow exerts an equal and opposite frictional force backward.

Understanding these laws helps explain why objects move (or stop moving) in predictable ways, allowing us to calculate distances and velocities in physics problems.

Potential Energy

Potential energy is energy stored due to the position of an object. Specifically, gravitational potential energy comes into play when the toboggan moves up the icy hill. Potential energy due to gravity can be calculated using the formula \( PE = mgh \), where:

• \( PE \) is the potential energy.
• \( m \) is the mass of the object.
• \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \)
• \( h \) is the vertical height above the starting point.

When the toboggan climbs the hill, its kinetic energy is transformed into potential energy until it comes to rest. The height reached (\( h \)) can be calculated once the entire kinetic energy has converted into potential energy, as shown in our solution.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. It is given by the formula \( KE = \frac{1}{2}mv^2 \), where:

• \( KE \) is the kinetic energy.
• \( m \) is the mass of the object.
• \( v \) is the velocity of the object.

In the exercise, kinetic energy considerations help us understand how fast the skier and toboggan are moving. During part (a), the skier's initial kinetic energy goes into overcoming kinetic friction. In part (b), part of the skier's initial kinetic energy remains as they reach the end of the patch, allowing us to calculate the final velocity. Meanwhile, for the toboggan in part (c), its complete kinetic energy converts into potential energy, allowing us to calculate the height it reaches. Understanding kinetic energy is crucial in analyzing motion problems in physics.