Comprehensive Analysis of Conservation of Mechanical Energy: The Skier Problem

Theoretical Framework: The Law of Conservation of Mechanical Energy

In the study of classical mechanics, the Law of Conservation of Mechanical Energy is a fundamental principle derived from the Work-Energy Theorem. This law states that in an isolated system where only conservative forces—such as gravity—perform work, the total mechanical energy ($E_{total}$) remain constant over time. Mechanical energy is defined as the sum of the kinetic energy ($KE$), which is the energy of motion, and the potential energy ($PE$), which is the energy stored due to an object's position within a gravitational field. For a system featuring a skier moving between two heights, the conservation law is mathematically expressed as follows:

$E_{initial} = E_{final}$

$KE_A + PE_A = KE_B + PE_B$

In this specific scenario, the transcript emphasizes that no friction was involved. Friction is a non-conservative force that converts mechanical energy into thermal energy (heat). Because friction is absent, there is no work done by non-conservative forces ($W_{nc} = 0$), ensuring that the total mechanical energy at Hill A is identical to the total mechanical energy at Hill B. The intervening path, such as the descent into a valley, does not affect the final value because gravity is a conservative force, meaning work done by gravity is independent of the path taken.

System Parameters and Constant Identification

To solve the kinematic problem presented, we must first document all given quantitative data and physical constants. The specific variables provided in the transcript are:

• Mass of the Skier ($m$): $67.5\,kg$
• Height of Hill A ($h_A$): $39\,m$
• Height of Hill B ($h_B$): $11\,m$
• Initial Speed at Top of Hill A ($v_A$): $12.6\,m/s$
• Acceleration due to Gravity ($g$): Approximated at standard Earth gravity, $9.8\,m/s^2$

It is important to note that while the mass of the skier is provided ($67.5\,kg$), in a frictionless environment where gravity is the only force doing work, the mass of the object will ultimately cancel out of the conservation equation. This is because both kinetic energy ($\frac{1}{2}mv^2$) and gravitational potential energy ($mgh$) are directly proportional to mass.

Defining Energetic Components: Kinetic and Potential Energy

The energy of the skier is divided into two discrete forms. Kinetic Energy ($KE$) represents the capacity of the skier to perform work by virtue of their velocity. The formula for kinetic energy is:

$KE = \frac{1}{2}mv^2$

Gravitational Potential Energy ($PE$) represents the energy an object possesses due to its position in a gravitational field relative to a reference point (usually the lowest point in the problem, such as the valley floor). The formula is:

$PE = mgh$

At Hill A, the skier possesses both initial kinetic energy (due to their speed of $12.6\,m/s$) and initial potential energy (due to their elevation of $39\,m$). As the skier descends into the valley, potential energy is converted into kinetic energy, reaching a maximum speed at the lowest point. As the skier ascends Hill B, kinetic energy is converted back into potential energy. The task is to determine the remaining kinetic energy (and thus velocity) at the height of $11\,m$.

Mathematical Derivation and Procedural Solution

To find the speed of the skier at the top of Hill B ($v_B$), we begin by setting up the full energy balance equation:

$\frac{1}{2}m(v_A)^2 + mgh_A = \frac{1}{2}m(v_B)^2 + mgh_B$

To simplify the calculation, we divide the entire equation by the mass ($m$), confirming that the motion is independent of the skier's weight:

$\frac{1}{2}(v_A)^2 + gh_A = \frac{1}{2}(v_B)^2 + gh_B$

Next, we isolate the variable for the final velocity squared ($v_B^2$) by multiplying the entire equation by $2$ and rearranging the terms:

$(v_B)^2 = (v_A)^2 + 2g(h_A - h_B)$

This specific form of the equation highlights that the final speed depends on the initial speed and the change in height (the vertical displacement $\Delta h = h_A - h_B$). Substituting the numerical values into the equation:

$(v_B)^2 = (12.6\,m/s)^2 + 2 \times 9.8\,m/s^2 \times (39\,m - 11\,m)$

$(v_B)^2 = 158.76\,m^2/s^2 + 19.6\,m/s^2 \times (28\,m)$

$(v_B)^2 = 158.76\,m^2/s^2 + 548.8\,m^2/s^2$

$(v_B)^2 = 707.56\,m^2/s^2$

To find the velocity $v_B$, we take the square root of both sides:

$v_B = \sqrt{707.56\,m^2/s^2}$

$v_B \approx 26.6\,m/s$

Summary of Results and Physical Implications

The skier's speed at the top of Hill B is calculated to be approximately $26.6\,m/s$. This result is logically consistent within the framework of physics: because Hill B ($11\,m$) is significantly lower than Hill A ($39\,m$), much of the initial potential energy has been permanently converted into kinetic energy, resulting in a final speed that is substantially higher than the initial speed of $12.6\,m/s$. In a real-world scenario, factors such as air resistance and friction between the skis and the snow would perform negative work, reducing the final speed. However, under the idealized conditions specified in the problem, the conservation of mechanical energy provides an exact prediction of the skier's kinematic state.