Problem 62 - College Physics

Question

\(\bullet\) \(\bullet\) A 62.0 kg skier is moving at 6.50 \(\mathrm{m} / \mathrm{s}\) on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 3.50 \(\mathrm{m}\) long. The coefficient of kinetic friction between this patch and her skis is \(0.300 .\) After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 \(\mathrm{m}\) high. (a) How fast is the skier moving when she gets to the bottom of the hill? (b) How much internal energy was generated in crossing the rough patch?

Step-by-Step Solution

Answer

(a) 8.74 m/s; (b) 637.77 J

Step 1: Calculate Initial Kinetic Energy

The initial kinetic energy \( K_i \) of the skier is given by the formula \( K = \frac{1}{2}m v^2 \), where \( m \) is the skier's mass and \( v \) is her velocity. Substituting the given values, \( K_i = \frac{1}{2} \times 62.0 \times (6.50)^2 \). Calculate to find \( K_i = 1308.25 \; \text{J} \).

Step 2: Calculate Work Done by Friction

The work done by the friction \( W_f \) is calculated using \( W_f = f_k \times d \), where \( f_k = \mu_k \times m \times g \) is the force of kinetic friction and \( d \) is the distance. \( f_k = 0.300 \times 62.0 \times 9.81 \), thus \( W_f = f_k \times 3.50 \). Substitute to get \( W_f = 637.77 \; \text{J} \).

Step 3: Calculate Skier's Speed After Rough Patch

The final kinetic energy right after crossing the patch is \( K_f = K_i - W_f \). So, \( K_f = 1308.25 - 637.77 \). Then, convert the final kinetic energy back to velocity using \( K_f = \frac{1}{2} m v^2 \). Solve \( v \) to get approximately \( 4.783 \; \text{m/s} \).

Step 4: Calculate Speed at the Bottom of the Hill

At the top of the hill, gravitational potential energy \( U = mgh \) is converted to kinetic energy. Before descending, \( U = 62.0 \times 9.81 \times 2.50 = 1520.25 \; \text{J} \). At the bottom, all potential energy becomes kinetic energy: \( K_{total} = K_f + U = \frac{1}{2}mv^2 \). Solve for \( v \), finding approximately \( 8.74 \; \text{m/s} \).

Step 5: Calculate Internal Energy Generated

The internal energy generated is equal to the work done by friction, which we found in Step 2. Therefore, the internal energy generated is \( 637.77 \; \text{J} \).

Key Concepts

Coefficient of Kinetic Friction

The coefficient of kinetic friction, often denoted as \( \mu_k \), is a crucial factor in understanding how surfaces interact when they are in motion relative to each other. It comprises the ratio of the force of kinetic friction to the normal force. Essentially, it tells us how "slippery" or "grippy" a surface is when an object moves across it.
For our skier, the coefficient \( \mu_k = 0.300 \) indicates a reasonably moderate resistance against her skis as she passes over the rough patch. It plays a pivotal role in calculating the work done by friction because frictional force \( f_k \) relies directly on this coefficient (\( f_k = \mu_k \times m \times g \)).
In everyday scenarios, different materials will have various coefficients, helping us predict how an object might slide or stick to another, which is important in designing anything from car tires to ski tracks.

Potential Energy

Potential energy is the energy an object has due to its position or state. For the skier, as she reaches the top of the hill, she has gravitational potential energy because she is elevated above ground level.
This energy is computed using the formula \( U = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is height. When the skier is at the top of the 2.50 m high hill, her potential energy is \( 1520.25 \; \text{J} \).
This stored energy has the potential to do work on the skier as she descends the hill, converting into kinetic energy that increases her speed. Understanding potential energy helps explain how position changes affect energy dynamics in physical systems, akin to having water in a dam ready to flow.

Work Done by Friction

Work done by friction is a concept that helps quantify how much energy is diverted from an object's kinetic energy as it moves across a rough surface. For our skier, the work done by friction \( W_f = 637.77 \; \text{J} \), is calculated by considering the force of friction \( f_k \) times the distance over which it acts (\( W_f = f_k \times d \)).
With her moving across a 3.5 m patch, the coefficient of kinetic friction determines how much energy is lost in overcoming the surface's resistance.
This work transforms some kinetic energy into internal energy, ultimately slowing her down. Such calculations are vital in disciplines like engineering and physics to evaluate energy losses in systems due to friction.

Energy Conversion

Energy conversion is central to physics, emphasizing the transformation of energy from one form to another. In skiing dynamics, it's fascinating to see potential and kinetic energy intimately linked via conversion processes.
As the skier moves, her initial speed gives her kinetic energy, which is partly converted to internal energy because of friction on the rough patch. When she moves downhill, her gravitational potential energy is reconverted into kinetic energy, aiding in her acceleration.
These transformations follow the law of conservation of energy, signifying energy doesn't disappear but merely changes form. Grasping this concept helps in various fields, analyzing anything from amusement park rides to electrical systems.

Skiing Dynamics

Skiing dynamics intricately involves the play between different forces and energies acting upon the skier. Factors such as gravitational force, frictional forces, kinetic energy, and potential energy dictate a skier’s speed and motion.
When the skier hits the rough patch, friction plays a significant role in reducing her speed due to energy loss, while descending the hill converts stored potential energy back into kinetic, increasing velocity.
Understanding skiing dynamics not only aids in improving skiing techniques but also assists in designing better skis and safer ski courses. It reveals how physics principles shine in sports, demonstrating tangible impacts of theoretical concepts on real-world applications.