Question

What coefficient of friction is required to stop a hockey puck sliding at \(12.5 \mathrm{~m} / \mathrm{s}\) initially over a distance of \(60.5 \mathrm{~m} ?\)

Short answer

Based on the given information (initial velocity of 12.5 m/s and stopping distance of 60.5 m), the coefficient of friction required to stop the hockey puck is approximately 0.131.

Step-by-step solution

Write down the information given

We are given: Initial velocity (u) = \(12.5 \mathrm{~m/s}\) Final velocity (v) = \(0 \mathrm{~m/s}\) (since the puck comes to a stop) Distance (s) = \(60.5 \mathrm{~m}\) We need to find: Coefficient of friction (μ)

Use the third equation of motion

We will use the third equation of motion: \(v^2 = u^2 + 2as\) We can rearrange the equation to solve for acceleration (a): \(a = \frac{v^2 - u^2}{2s}\) Now, plug the values into the equation: \(a = \frac{(0 \mathrm{~m/s})^2 - (12.5 \mathrm{~m/s})^2}{2(60.5 \mathrm{~m})}\) Calculate the value of a: \(a = \frac{-156.25 \mathrm{~m^2/s^2}}{121 \mathrm{~m}} = -1.29 \mathrm{~m/s^2}\) The negative sign indicates that the acceleration is acting in the opposite direction to the initial velocity, which is expected since it's a decelerating force.

Relate the acceleration to frictional force and the coefficient of friction

We know that \(F_\text{friction} = \mu * F_\text{normal}\), and the frictional force provides the deceleration. The normal force acting on the puck is equal in magnitude to its weight (mass x gravitational acceleration) since it's sliding horizontally. Therefore, \(F_\text{friction} = ma\) where m is the mass of the puck Now we can write: \(\mu * mg = ma\) Since the mass of the puck (m) is present in both sides of the equation, we can eliminate it: \(\mu * g = a\) Now, we can rearrange the equation to solve for the coefficient of friction (μ): \(\mu = \frac{a}{g}\)

Calculate the coefficient of friction

We have the values of a and g (gravitational acceleration, approx. 9.81 m/s²), so we can plug them into the equation: \(\mu = \frac{-1.29 \mathrm{~m/s^2}}{9.81 \mathrm{~m/s^2}}\) Calculate the value of μ: \(\mu \approx 0.131\) The coefficient of friction required to stop the hockey puck sliding initially at 12.5 m/s over a distance of 60.5 meters is approximately 0.131.

Key concepts

Third Equation of Motion

Understanding the third equation of motion is pivotal when we're analyzing objects in motion, especially during acceleration or deceleration. This equation, expressed as \(v^2 = u^2 + 2as\), relates the final velocity (\(v\)) of an object to its initial velocity (\(u\)), the acceleration (\(a\)), and the distance covered (\(s\)). In the case of a hockey puck sliding to a stop, we use this equation to determine the negative acceleration, or deceleration, that the puck experiences as friction works to stop it over the given distance. It's important to note the negative sign in the answer reflects deceleration—the puck is slowing down.

Frictional Force

Frictional force is the resisting force encountered by an object when it moves or attempts to move across a surface. It plays a crucial role in stopping moving objects and is calculated by the equation \(F_{\text{friction}} = \text{μ} \times F_{\text{normal}}\), where \text{μ} represents the coefficient of friction and \(F_{\text{normal}}\) represents the normal force. In our hockey puck scenario, the frictional force is what causes the puck to decelerate and eventually come to a halt. It is essential to differentiate between static and kinetic (sliding) friction—the former acting when the object is at rest, and the latter when the object is in motion.

Normal Force

The normal force is an essential component when calculating frictional forces. It is the force exerted by a surface to support the weight of an object resting on it, acting perpendicular to the surface. In many situations, like that of the hockey puck sliding horizontally, the normal force is equal to the object's weight, which is the mass (\(m\)) times the gravitational acceleration (\(g\)). Thus, on level surfaces, \(F_{\text{normal}} = m \times g\). The normal force is directly proportional to frictional force—the greater the normal force, the larger the frictional force at play.

Gravitational Acceleration

Gravitational acceleration, denoted by \(g\), is the acceleration due to gravity that acts on any object in free fall near the Earth's surface. Typically, its value is approximately \(9.81 \text{m/s}^2\). This value can vary slightly depending on altitude and location. In physics problems, gravitational acceleration allows us to calculate various forces, including weight and normal force. As seen in our example, it helps to relate the deceleration of the hockey puck to the frictional force acting against it.

Deceleration

Deceleration is the rate at which an object slows down. It's essentially negative acceleration and occurs when the final velocity of an object is less than its initial velocity. In the formula \(a = \frac{v^2 - u^2}{2s}\), a deceleration will have a negative value, indicating a reduction in speed. Understanding deceleration is critical when trying to determine the distance needed for an object to come to a complete stop, as well as the forces involved in this process, such as in our hockey puck example.