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A 60.0 -kg person, running horizontally with a velocity of \(+3.80 \mathrm{m} / \mathrm{s}\), jumps onto a \(12.0-\mathrm{kg}\) sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast \(30.0 \mathrm{m}\) on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?

Short Answer

Expert verified
(a) 3.17 m/s (b) 0.0171

Step by step solution

01

Understand the Conservation of Momentum

For part (a), we'll use the principle of conservation of momentum which states that the total momentum before the collision must equal the total momentum after the collision. Initially, the person is moving and the sled is at rest, thus:\[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \]where \( m_1 = 60.0 \text{ kg} \) is the mass of the person, \( v_1 = 3.80 \text{ m/s} \) is the initial velocity of the person, \( m_2 = 12.0 \text{ kg} \) is the mass of the sled, \( v_2 = 0 \text{ m/s} \) is the initial velocity of the sled, and \( v_f \) is the final velocity of the sled and person together.
02

Calculate Final Velocity After Collision

Substitute the values into the conservation of momentum equation:\[ (60.0 \times 3.80) + (12.0 \times 0) = (60.0 + 12.0) v_f \]This simplifies to:\[ 228 = 72 v_f \]Solve for \( v_f \):\[ v_f = \frac{228}{72} \approx 3.17 \text{ m/s} \]So, the final velocity after the collision is approximately \( 3.17 \text{ m/s} \).
03

Analyze Motion to Find Coefficient of Friction

For part (b), we'll use the fact that the sled and person coast to a stop over \( 30.0 \text{ m} \). We'll start by finding the acceleration using the formula:\[ v^2 = u^2 + 2as \]where \( v = 0 \text{ m/s} \) (final velocity), \( u = 3.17 \text{ m/s} \) (initial velocity), and \( s = 30.0 \text{ m} \) (distance). Substituting:\[ 0 = (3.17)^2 + 2a(30.0) \]\[ 0 = 10.05 + 60a \]Solving for \( a \), we find:\[ a = -\frac{10.05}{60} = -0.1675 \text{ m/s}^2 \]
04

Calculate Coefficient of Kinetic Friction

The force of kinetic friction \( f_k \) is given by:\[ f_k = \mu_k F_n \]where \( \mu_k \) is the coefficient of kinetic friction and \( F_n \) is the normal force (equal to the gravitational force \( mg \) here).Using Newton's second law, \( f_k = ma \):\[ \mu_k mg = ma \]\[ \mu_k = \frac{a}{g} \]Substituting the known values, \( a = -0.1675 \text{ m/s}^2 \) and \( g = 9.8 \text{ m/s}^2 \):\[ \mu_k = \frac{-0.1675}{9.8} \approx 0.0171 \]Thus, the coefficient of kinetic friction is approximately \( 0.0171 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coefficient of Kinetic Friction
When an object moves over a surface, it tends to experience a force that opposes its motion. This resisting force is called friction, and more specifically for moving objects, it is termed **kinetic friction**. The coefficient of kinetic friction, denoted by \( \mu_k \), is a value that represents the frictional force as a proportion of the normal force. It is unique to the combination of surface materials in contact.

To find \( \mu_k \) in mechanics problems like the one in the exercise, you can rearrange the following equation involving Newton's second law:
  • The force of kinetic friction: \( f_k = \mu_k F_n \)
  • \( F_n \) is the normal force, calculated as \( mg \) for horizontal surfaces.
By knowing the deceleration due to friction (found using kinematic equations), you can substitute it into:
  • \( ma = \mu_k mg \)
  • Simplifying gives \( \mu_k = \frac{a}{g} \)
Remember, having a low coefficient, like \( 0.0171 \) in the problem, indicates a very smooth interaction with little resistance.
Newton's Second Law
Newton's Second Law is one of the cornerstone principles of physics, providing a quantitative relationship between force, mass, and acceleration. It is concisely expressed by the formula \( F = ma \).

This formula indicates that the force acting on an object is equal to the mass of the object multiplied by the acceleration it experiences. It implies a few key points:
  • Force is directly proportional to both mass and acceleration.
  • A heavier object requires more force to achieve the same acceleration as a lighter object.
  • Acceleration is inversely proportional to mass when a constant force is applied.
In practical terms, if you need to find either the force required to accelerate an object or the acceleration resulting from a certain force, applying this law can provide answers directly or when rearranged as needed. In the problem scenario, we use this principle to relate the frictional stopping force to the acceleration experienced by the person and sled.
Conservation of Energy
The principle of conservation of energy tells us that in a closed system, the total energy remains constant. Energy can neither be created nor destroyed; it can only be transformed or transferred.

Although not directly used in the exercise, understanding conservation of energy helps analyze situations where mechanical energy becomes kinetic or potential, or dissipates through friction as thermal energy.
  • Kinetic Energy: Energy of motion, calculated as \( \frac{1}{2}mv^2 \)
  • Potential Energy: Stored energy related to an object's position, such as gravitational potential energy \( mgh \)
  • Work Done by Friction: Converts some kinetic energy into heat, slowing objects down.
In broader applications, while momentum being conserved enables us to handle the direct collision aspect of this problem, energy conservation can offer insights into the saws and transformations occurring, deepening our understanding of the entire motion process.

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