91Ó°ÊÓ

Kinematic equations are used to describe motion, especially when an object experiences constant acceleration. These equations relate key parameters of motion such as initial and final velocity, acceleration, time, and displacement.
For the hockey puck problem, we need to calculate the acceleration caused by the frictional force as it brings the puck to a stop over a distance of 15 meters.
The relevant kinematic equation is:

• \(v_f^2 = v_i^2 + 2ad\)

where:\(v_f\) is the final velocity and is 0 m/s because the puck stops. \(v_i\) is the initial velocity of 6.0 m/s, \(a\) is the acceleration, and \(d\) is the distance of 15 m.
By rearranging this equation, substituting the known values, and solving for \(a\), we determine that the acceleration is \(-1.2 \text{ m/s}^2\). The negative sign reflects that the force is opposing the motion. This equation is an essential tool to find the required acceleration given the other parameters of motion.

Newton's Second Law explains how the motion of an object changes in response to an applied force. This law is succinctly represented by the formula \( F = ma \), where \(F\) represents the force applied to the object, \(m\) is its mass, and \(a\) is the acceleration.
It reveals that force is directly proportional to both mass and acceleration.
In our exercise, once we found the puck's acceleration using the kinematic equations, we applied Newton's Second Law to find the frictional force.

• The mass \(m\) of the puck is 0.11 kg.
• The acceleration \(a\) calculated is \(-1.2 \text{ m/s}^2\), again negative due to the opposing direction.
• Thus, the frictional force \(f_f\) was calculated as \(0.132 \text{ N}\).

This force counteracts the puck's movement, highlighting how friction works to slow moving objects.

The coefficient of friction, denoted by \(\mu\), is a dimensionless scalar value that describes the ratio of the frictional force between two bodies and the force pressing them together. It tells us how much grip or resistance exists between surfaces.

• To find \(\mu\), we use the equation \(f_f = \mu mg\).
• Here, \(f_f\) is the frictional force \(0.132 \text{ N}\),
• \(m\) is the mass of the puck \(0.11 \text{ kg}\),
• and \(g\) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\).

Substituting these into the formula, \(\mu = \frac{0.132}{0.11 \times 9.8} \approx 0.122\).
This coefficient indicates how the ice's surface resists the sliding puck. A lower value like 0.122 suggests a slick surface, which is typical of ice.