Question

On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 m/s encounters a rough patch that reduces her speed to 1.65 m/s due to a friction force that is 25% of her weight. Use the work\(-\)energy theorem to find the length of this rough patch.

Short answer

The length of the rough patch is approximately 1.43 meters.

Step-by-step solution

Identify the Work-Energy Theorem

The work-energy theorem states that the work done by the forces on an object equals the change in its kinetic energy. Mathematically, it is expressed as \( W = \Delta KE \). Here, \( W \) is the work done by friction, and \( \Delta KE \) is the change in kinetic energy of the skater.

Calculate Initial and Final Kinetic Energy

The initial kinetic energy (\( KE_i \)) is calculated using \( KE_i = \frac{1}{2} m v_i^2 \), where \( m \) is the skater's mass and \( v_i \) is the initial velocity (3.0 m/s). The final kinetic energy (\( KE_f \)) is \( KE_f = \frac{1}{2} m v_f^2 \), where \( v_f \) is the final velocity (1.65 m/s). Note that \( m \) can be factored out since it's the same for both states.

Calculate Change in Kinetic Energy

The change in kinetic energy (\( \Delta KE \)) is \( \Delta KE = KE_f - KE_i \). Substitute the values for initial and final kinetic energy, resulting in \( \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) \).

Express Work Done by Friction

The work done by friction (\( W \)) is given by \( W = f \times d \), where \( f \) is the friction force and \( d \) is the distance of the rough patch (unknown). The friction force \( f \) is 25% of the skater's weight, so \( f = 0.25mg \), where \( g \) is the acceleration due to gravity (9.8 m/s²).

Set up the Equation and Solve for Distance

Using the relation from the work-energy theorem \( W = \Delta KE \) and substituting \( f \) and \( W \) in, we have \( 0.25mgd = \frac{1}{2} m (v_f^2 - v_i^2) \). Here, \( m \) cancels out. Solve the equation for \( d \): \[ d = \frac{(v_i^2 - v_f^2)}{2 \times 0.25 \times g} \] Substitute \( v_i = 3.0 \) m/s, \( v_f = 1.65 \) m/s, and \( g = 9.8 \) m/s² into the equation to find \( d \).

Calculate the Value of d

Plug in the values: \( d = \frac{(3.0^2 - 1.65^2)}{2 \times 0.25 \times 9.8} \) Calculate to find \( d \approx 1.43 \) meters.

Key concepts

Kinetic Energy

Kinetic energy is an essential concept to understand when analyzing motion. It is the energy that an object possesses due to its motion, calculated using the formula: \[ KE = \frac{1}{2} m v^2 \] where:

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

In this exercise, the skater's kinetic energy changes as they move across the rough patch of the ice rink.
Initially, the skater moves with a velocity of 3.0 m/s. As she reaches the rough patch, her velocity decreases to 1.65 m/s due to friction. This reduction in speed means a drop in her kinetic energy. The difference between the initial and final kinetic energy is used to determine the work done by friction, since:\[ \Delta KE = KE_f - KE_i \]Understanding kinetic energy is vital, as it helps us calculate the necessary work done to alter an object's speed, providing insight into the forces at play.

Friction Force

Friction is the force that opposes the motion of one surface sliding over another. It is a crucial factor when considering the movement on surfaces like ice rinks. In the given scenario, even though the ice surface is mostly frictionless, there exists a rough patch.
The friction force in this example is calculated as a fraction of the skater's weight. Specifically, it's 25% of her weight: \[ f = 0.25mg \]where:

• \( f \) is the friction force,
• \( m \) is the mass of the skater, and
• \( g \) is the acceleration due to gravity (9.8 m/s²).

This friction force slows down the skater, changing her kinetic energy. The work done by this force is equal to the friction force times the distance (\( d \)) over which it acts:\[ W = f \times d \]The frictional work computed is then set equal to the change in kinetic energy as per the work-energy theorem, illustrating how friction affects movement and energy changes on the ice.

Acceleration Due to Gravity

Acceleration due to gravity, represented as \( g \), is a crucial constant in physics. It is approximately 9.8 m/s² on Earth. This value is paramount when calculating forces and motion, especially when considering weights and the effects of gravity.
In this exercise, the acceleration due to gravity is employed in determining the friction force that acts on the skater. Since friction is 25% of her weight, and weight is calculated as mass times gravitational acceleration (\( mg \)), \( g \) is a key component of the friction force equation:\[ f = 0.25mg \]Understanding \( g \) enables us to comprehend how different forces interplay when an object is on Earth, affecting everything from simple motions to more complex energy exchange processes like those experienced by the skater as she encounters the rough patch on the ice rink. Such fundamental concepts help us model real-world scenarios effectively and imagine how objects behave under the influence of gravity.