Problem 23 - College Physics

Question

\(\bullet\) A 61 kg skier on level snow coasts 184 m to a stop from a speed of 12.0 \(\mathrm{m} / \mathrm{s}\) . (a) Use the work-energy principle to find the coefficient of kinetic friction between the skis and the snow. (b) Suppose a 75 \(\mathrm{kg}\) skier with twice the starting speed coasted the same distance before stopping. Find the coefficient of kinetic friction between that skier's skis and the snow.

Step-by-Step Solution

Answer

(a) \( \mu_k = 0.039 \); (b) \( \mu_k = 0.163 \)

Step 1: Understanding the Problem

We need to compute the coefficient of kinetic friction (\( \mu_k \)) for two different cases using the work-energy principle. The principle states that the work done by friction is equal to the change in kinetic energy of the skier.

Step 2: Gathering Known Variables for Part (a)

For the 61 kg skier:- Mass \( m = 61 \ \text{kg} \)- Initial velocity \( v_i = 12 \ \text{m/s} \)- Final velocity \( v_f = 0 \ \text{m/s} \)- Distance \( d = 184 \ \text{m} \)- Gravitational acceleration \( g = 9.8 \ \text{m/s}^2 \)

Step 3: Calculating the Change in Kinetic Energy for Part (a)

The initial kinetic energy \( KE_i = \frac{1}{2}mv_i^2 \) and the final kinetic energy \( KE_f = 0 \).Thus, the change in kinetic energy is \( \Delta KE = KE_f - KE_i = 0 - \frac{1}{2}(61 \ kg)(12 \ \text{m/s})^2 \).Calculate \( \Delta KE = -4392 \ \text{J} \).

Step 4: Finding Work Done by Friction for Part (a)

The work done by friction (\( W_f \)) is \( W_f = -\Delta KE \).This means the work done by friction is \( 4392 \ \text{J} \).

Step 5: Relating Work Done by Friction to Frictional Force for Part (a)

Work done by friction can also be expressed as \( W_f = f_k \cdot d \), where \( f_k = \mu_k mg \).Thus, \( 4392 \ \text{J} = \mu_k \times 61 \ \text{kg} \times 9.8 \ \text{m/s}^2 \times 184 \ \text{m} \).Solve for \( \mu_k \).

Step 6: Solution to Part (a)

Calculate \( \mu_k = \frac{4392 \ \text{J}}{61 \ \text{kg} \times 9.8 \ \text{m/s}^2 \times 184 \ \text{m}} = 0.039 \).

Step 7: Setup Known Variables for Part (b)

For the 75 kg skier with twice the initial speed:- Mass \( m = 75 \ \text{kg} \)- Initial velocity \( v_i = 24 \ \text{m/s} \) (double the initial speed)- Distance \( d = 184 \ \text{m} \) (same as before)- Final velocity \( v_f = 0 \text{ m/s} \)

Step 8: Calculate Change in Kinetic Energy for Part (b)

Initial kinetic energy \( KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(75 \ kg)(24 \ \text{m/s})^2 \).Calculate \( \Delta KE = 0 - KE_i = -21600 \ \text{J} \).

Step 9: Find Work Done by Friction for Part (b)

The work done by friction for Part (b) is equal to the change in kinetic energy: \( 21600 \ \text{J} \).

Step 10: Relate Work Done to Frictional Force for Part (b)

Use the equation \( W_f = \mu_k \times m \times g \times d \).Substitute known values: \( 21600 \ \text{J} = \mu_k \times 75 \ \text{kg} \times 9.8 \ \text{m/s}^2 \times 184 \ \text{m} \).Solve for \( \mu_k \).

Step 11: Solution to Part (b)

Calculate \( \mu_k = \frac{21600 \ \text{J}}{75 \ \text{kg} \times 9.8 \ \text{m/s}^2 \times 184 \ \text{m}} = 0.163 \).

Key Concepts

Kinetic Friction

Kinetic friction is a force that opposes the relative sliding motion between two surfaces in contact. In the context of a skier moving on snow, kinetic friction is the resistance that the snow exerts on the skis, slowing the skier's motion. The magnitude of this force can be determined by the coefficient of kinetic friction (\( \mu_k \)), which is a dimensionless number indicating how slippery or sticky the surfaces are.

• High \( \mu_k \) means more friction
• Low \( \mu_k \) means less friction

To calculate \( \mu_k \), the work-energy principle is used:

• The work done by the frictional force is equal to the change in kinetic energy of the skier.
• This work is calculated as the product of the frictional force, the distance the skier travels until they stop, and is expressed as \( W_f = f_k \cdot d \).

By rearranging the work formula \( f_k = \mu_k m g \), the coefficient can be isolated to find \( \mu_k \). In our exercise, different values of \( \mu_k \) were calculated for skiers with different masses and initial speeds.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. This type of energy depends on both the mass (\( m \)) of the object and its velocity (\( v \)). It is calculated using the formula:\[ KE = \frac{1}{2} m v^2 \]For our skiers, initial kinetic energy can be substantial considering their relatively high starting speeds. As the skier moves across the snow and comes to a stop, this energy decreases to zero. The difference in kinetic energy represents the work done by friction.

• If the skier's mass or speed increases, the kinetic energy rises.
• Halving the speed results in a quarter of the original kinetic energy.
• Doubling the mass results in twice the original kinetic energy.

Understanding the relationship between kinetic energy and changes in motion helps one grasp why friction, acting over the skiing distance, effectively stops the skier.

Skier Motion

When a skier moves across snow, various forces and energies come into play. Initially, the skier begins with kinetic energy due to their velocity. As they glide forward:

• Kinetic friction from the snow surface acts against the direction of motion.
• This friction gradually reduces the skier's speed until they eventually come to a stop.

The uniform deceleration caused by friction depends on multiple factors, including the skier's mass, the initial speed, and the nature of the ski-snow contact — expressed through the coefficient of kinetic friction.
In scenarios like the one we explored, the work-energy principle allows us to quantify these interactions and predict outcomes such as:

• The distance required to come to a halt for different speeds and masses.
• The variations in friction needed to achieve a stop under different conditions.

These calculations break down the complex, real-world motion into understandable parts, illustrating how energy principles govern everyday activities like skiing.