Question

Santa's reindeer pull his sleigh through the snow at a speed of \(2.561 \mathrm{~m} / \mathrm{s}\) The mass of the sleigh, including Santa and the presents, is \(540.3 \mathrm{~kg} .\) Assuming that the reindeer can provide a total power of \(2.666 \mathrm{hp}\), what is the coefficient of friction between the runners of the sleigh and the snow?

Short answer

Answer: The coefficient of friction between the runners of Santa's sleigh and the snow is approximately 0.1467.

Step-by-step solution

Find the friction force

Using the formula, Power = Force × Velocity, we can find the friction force acting on the sleigh: \(Force = \frac{Power}{Velocity}\) Given, Power (P) = 2.666 hp (we need to convert hp to watts, 1 hp = 746 watts) and Velocity (v) = 2.561 m/s. Now substituting the values: \(Force = \frac{2.666 \times 746}{2.561}\) \(Force = \frac{1991.436}{2.561}\) \(Force \approx 777.145 \mathrm{N}\)

Calculate Normal force (Fn)

The normal force (Fn) is equal to the gravitational force acting on the sleigh (Fg), which can be calculated as Fn = Fg = Mass(m) × Gravitational acceleration(g). Given: Mass (m) = 540.3 kg Gravitational acceleration (g) = 9.8 m/s² Fn = m × g Fn = 540.3 × 9.8 Fn ≈ 5294.94 N

Calculate Coefficient of friction (μ)

Now, we can use the formula of friction force (Ff) to find the coefficient of friction (μ): Friction force (Ff) = μ × Fn Here, Friction force (Ff) = 777.145 N, Normal force (Fn) ≈ 5294.94 N. Solving for μ: \(\mu = \frac{Ff}{Fn}\) \(\mu = \frac{777.145}{5294.94}\) \(\mu \approx 0.1467\) The coefficient of friction between the runners of the sleigh and the snow is approximately 0.1467.

Key concepts

Physics Problem Solving

Mastering physics problem solving involves translation of a real-world scenario into a mathematical framework. Let's take the example of Santa's reindeer sleigh. To deduce the coefficient of friction, we must first understand the forces at play. Friction force, a concept central to this problem, opposes the motion of the sleigh. To solve this, we carefully set up equations based on known physics principles, such as the relationship between power, force, and velocity.

Ensuring that units are consistent, in this case converting horsepower to watts, is crucial. With precise calculation steps, we derived the frictional and normal forces, which ultimately allowed us to find the coefficient of friction. Paying attention to the step-by-step solution process helps in grasping not only how to apply formulas but also in understanding the underlying physics concepts involved.

Friction Force Calculation

Friction force plays a significant role in how objects move on surfaces. To calculate the friction force acting on Santa's sleigh, we used the formula where Power equals Force multiplied by Velocity.

Given the power in horsepower, it's pivotal to convert it to watts, the standard unit of power in the International System of Units (SI), before plugging the values into the equation. This ensures we're using consistent units to find the force in newtons. The knowledge of converting different units and applying relevant formulas correctly is invaluable in calculating friction forces in various physics problems.

Power to Force Conversion

In converting power to force, we encounter the formula: Power = Force × Velocity. This is an integral concept in physics that illustrates the interplay between an object's movement and the energy transferred. Power, measured in watts, quantifies how quickly work is done, while force, measured in newtons, is the push or pull on an object that can cause it to accelerate.

When we solve for force—given the power output of the reindeer and the velocity of Santa's sleigh—we unlock the first step in understanding the dynamics of the sleigh's motion. Through this conversion, we can understand the amount of effort, or force, the reindeer are exerting against the snow's resistance, thus translating a unit of power into a tangible force that can be further analyzed in different physics applications.