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Two figure skaters, one weighing \(625 \mathrm{~N}\) and the other \(725 \mathrm{~N}\) push off against cach other on frictionless ice. (a) If the heavicr skater travels at \(1.50 \mathrm{~m} / \mathrm{s}\), how fast will the lighter one travel? (b) How much kinetic energy is "created" during the skaters' maneuver, and where does this cnergy come from?

Short Answer

Expert verified
The speeds of the skaters after the push-off are calculated to be \(V_{H_f} = 1.50 m/s\) and \(V_{L_f} = 1.74 \, m/s\). The kinetic energy is created by the work done by the skaters pushing off against each other.

Step by step solution

01

Define the variables

Let's denote the initial weights of the heavier and lighter skates as \(W_H = 725 N\) and \(W_L = 625 N\) respectively, their final velocities as \(V_{H_f} = 1.50 m/s\) and \(V_{L_f}\) respectively (which we want to find), and the initial velocities as \(V_{H_i} = 0 m/s\) and \(V_{L_i} = 0 m/s\) since they were stationary before pushing off each other.
02

Use the conservation of momentum to find \(V_{L_f}\)

From the law of conservation of momentum, the momentum of the system before the skaters push off each other is equal to the momentum of the system after they push off. That is, \(W_H \cdot V_{H_i} + W_L \cdot V_{L_i} = W_H \cdot V_{H_f} + W_L \cdot V_{L_f}\). Since \(V_{H_i}\) and \(V_{L_i}\) are zero, we have \(W_H \cdot V_{H_f} = W_L \cdot V_{L_f}\), that gives us \(V_{L_f} = W_H \cdot V_{H_f} / W_L\). Substituting given values, \(V_{L_f} = 725 N \cdot 1.50 m/s / 625 N\).
03

Calculate the kinetic energy

The kinetic energy (\( K.E. \)) of an object is given by \( K.E. = 0.5 * m * v^2 \) where \( m \) is the mass and \( v \) the velocity of the object. However, since we have the weights, not the masses of the skaters, we can use the relationship that Newton's weight is equal to mass times gravity. Therefore, \( m = W/g \), where \( g \) is the gravitational acceleration with a value of \( 9.81 \, m/s^2 \). Now let's calculate the total kinetic energy before and after the skaters push off. Before the push off, both skaters are stationary, so the total kinetic energy is zero. After the push off, the total energy \( K.E. = 0.5 * W_H/g * V_{H_f}^2 + 0.5 * W_L/g * V_{L_f}^2 \).
04

Find out where the kinetic energy comes from

The kinetic energy comes from the work done by the skaters when they push off against each other. The work done by each skater becomes the kinetic energy of the other skater. This work is what makes the skaters move; it converts into kinetic energy as they start sliding across the ice with their new respective velocities.

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Key Concepts

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

Kinetic Energy
Kinetic energy is an essential concept in physics representing the energy an object possesses due to its motion. It's given by the formula \( K.E. = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity of the object. In our exercise, the two skaters transform their exerted power and energy into motion when they push off against each other.
The heavier skater, weighing 725 N, and the lighter skater, weighing 625 N, were initially stationary. As they push off, each generates kinetic energy from the work they exerted. This kinetic energy wasn't created from nowhere—it resulted from the force applied by each skater pushing off from the other, adhering to the conservation of energy principle. Before they moved, each had zero kinetic energy, but after pushing off, they both acquired kinetic energy determined by their masses and final speeds.
The relationship between weight and mass in this scenario is essential. Since weight is the force due to gravity on an object, we can convert it to mass using Newton's second law: \( m = \frac{W}{g} \), where \( g = 9.81 \ m/s^2 \). By determining each skater's mass, we can calculate their kinetic energy once their velocities are known.
Frictionless Surface
Understanding a frictionless surface is crucial when discussing momentum and energy conservation exercises like the skater problem. A frictionless surface implies no opposing force acts to slow or stop an object's motion. In reality, such surfaces don't truly exist, but for theoretical problems, they help simplify calculations by eliminating the variable of friction.
On a frictionless surface, once in motion, objects continue moving indefinitely in the absence of external forces, aligning with Newton's First Law of Motion. In our skater scenario, this means that after pushing off, the skaters don't slow down due to friction—they maintain their velocities until another force acts on them.
This concept of a frictionless surface helps isolate other variables in physics problems, allowing learners to focus on fundamental forces and calculations like momentum and kinetic energy without the added complexity of friction.
Newton's Laws
Newton's Laws of Motion are foundational in understanding how objects interact and move in our world. They also play a crucial role in the problem involving skaters on ice. Let's explore how each law connects to the scenario:
  • First Law (Inertia): An object at rest stays at rest unless acted upon by an external force. Initially, both skaters are stationary, but once they push off, the external force (their mutual push) sets them in motion.
  • Second Law (F = ma): This law explains how forces result in acceleration. The force exerted during their push-off sets each skater into motion with a certain acceleration, depending on their mass.
  • Third Law (Action-Reaction): Newton's Third Law is most directly witnessed here. When the skaters push against each other, the action of one skater is met with an equal and opposite reaction, propelling both in opposite directions.
In essence, these laws provide the framework to understand how the skaters' movements generate kinetic energy and adhere to the conservation of momentum in the absence of friction.

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Most popular questions from this chapter

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