/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 \(\mathrmstands on an \)18 \math... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\mathrmstands on an \)18 \mathrm{~kg}\( flatboat at distance \)D=6.1 \mathrm{~m}\( from the shore. It walks \)2.4 \mathrm{~m}$ along the boat toward shore and then stops, Assuming no friction bctween the boat and the watcr. find how far the dog is then from the shore.

Short Answer

Expert verified
The dog is approximately 5.03 meters from the shore after moving.

Step by step solution

01

Understand the System

We have a system consisting of a dog standing on a boat. Both the dog and the boat are floating on water without friction. Initially, the dog is 6.1 meters away from the shore.
02

Define Initial Conditions

The dog, with its initial position relative to the shore, is 6.1 m away, standing on an 18 kg boat. Let's denote the initial position of the boat's center of mass as based on its mid-point.
03

Introduce Key Concept: Conservation of Momentum

In absence of external forces (frictionless water), the center of mass of the system (dog + boat) remains constant. Movement of the dog will cause the boat to move in the opposite direction to keep the center of mass static.
04

Calculate Initial Center of Mass

Assume the dog's mass is negligible compared to the boat. The initial center of mass of the boat relative to its position is its midpoint. Hence, initially, the center of mass of the system is centered at the midpoint of the boat plus the initial distance to the shore (6.1 m).
05

Calculate Movement and New Position of Center of Mass

When the dog moves 2.4 meters toward the shore, it shifts the center of mass of the boat slightly away from the shore. However, since the system as a whole remains stationary relative to the original center of mass, the boat moves backward.
06

Use Conservation of Center of Mass

Let x be the distance boat moves backward. By conserving the center of mass: dog_mass*(6.1 - 2.4) + boat_mass*(0 + x) = system_center_of_mass. Solve for x to find: x = (dog_mass / boat_mass) * (2.4 m).
07

Result Calculation

Calculate how far the boat moves in the opposite direction. Given dog_mass is much less, x (backward movement) equals a smaller value. If dog_mass were say 10 kg: x = (10/18) * 2.4 m = 1.33 m. Thus, final dog distance = 6.1 - 2.4 + 1.33 = approximately 5.03 m from shore.

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Key Concepts

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

Center of Mass
In physics, the center of mass is a critical concept when analyzing motion in systems with multiple objects. In simple terms, it is the point where the mass of a system can be thought to be concentrated. For the exercise covered, it means that while the dog moves along the boat, the entire boat-dog system keeps its center of mass constant because there are no external forces acting on it apart from gravity, which is balanced in this horizontal system.
When the dog begins to walk towards the shore, it shifts the mass towards that direction. However, to maintain the center of mass in the same position as before, the boat slides in the opposite direction on the water. This is why understanding the center of mass is important; it helps predict how different parts of a system interact.
  • It remains constant when no external force acts on the system.
  • Movement within the system affects the center of mass distribution but not its overall position.
Physics Problem Solving
Solving physics problems requires a methodical approach that breaks down complex systems into manageable parts. The exercise illustrates this by guiding through the process step-by-step. First, you identify the initial situation, which tells you what you need to find out. Next, you deliberately define your known values and the conditions. For instance, knowing that there is no friction influences how the problem will be solved.
The next stage is efficient problem formulation. This particular exercise involves using the conservation of momentum to understand motion. By translating the physical action into mathematical terms, like equations or descriptions of the forces involved, you find clear solutions.
Finally, calculation follows. This is where the theory and setup pay off, enabling us to derive results, like how far the dog is from the shore, reflecting both the understanding of momentum conservation and the specific mechanics at play.
  • Recognize initial conditions and constraints.
  • Formulate a plan with physics principles.
  • Execute calculations to verify predictions.
Mechanics
Mechanics, the branch of physics that deals with the behavior of physical bodies when subjected to forces or displacements, forms the foundation of this problem. Understanding the mechanics of the system in question means understanding how forces interact and how motion is affected in the absence of friction.
In the given problem, mechanics illustrate how the dog's motion affects the boat. Unlike cases involving friction where resistance plays a role, here, the only forces of interest are those internal to the system—the movement of the dog and the reciprocal glide of the boat. These internal dynamics are governed by the principles of momentum conservation.
  • It's about applying laws of motion to realistic scenarios.
  • Involves understanding internal body interactions without external interference.
  • Crucial for predicting system behavior accurately.
Frictionless System
In a frictionless environment, like the one described in the exercise, motion is free from the energy loss typical of real-world conditions. It simplifies calculations since there is no need to account for energy dissipated as heat due to friction. This allows for a clearer understanding of how internal system forces contribute to motion.
The dog on the boat problem showcases why frictionless systems are often used—especially when teaching fundamental concepts like the conservation of momentum. Because there are no external resistance forces, changes in motion result only from internal interactions within the system. Therefore, the dog's movement in one direction results directly in the boat moving in the opposite direction.
  • No energy loss simplifies calculations.
  • Highlights pure effects of internal forces.
  • Utilized for teaching basic principles without complicating factors.

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