Ch. 05 - Using Newton's Laws: Friction, Circular Motion, Drag Forces

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 05 - Using Newton's Laws: Friction, Circular Motion, Drag Forces

Problem 104c

Chapter 5, Problem 104c

A 68-kg water skier is being accelerated by a ski boat on a flat ('glassy') lake. The coefficient of kinetic friction between the skier's skis and the water surface is μₖ = 0.25 (Fig. 5–59). Explain why the skier's acceleration in part (b) is greater than that in part (a).

Verified step by step guidance

Step 1: Begin by understanding the forces acting on the skier. The skier experiences a forward force due to the tension in the rope (applied by the boat) and a backward force due to kinetic friction. The net force determines the skier's acceleration according to Newton's second law:

F

_net = ma.

Step 2: The force of kinetic friction is given by the formula:

F

_friction = μₖmg, where

μ

_k is the coefficient of kinetic friction,

m

is the mass of the skier, and

g

is the acceleration due to gravity.

Step 3: In part (b), the skier's acceleration is greater because the net force acting on the skier is larger. This could be due to an increase in the tension force applied by the boat, which overcomes the frictional force more effectively. The net force is calculated as:

F

_net = F_tension - F_friction.

Step 4: Compare the scenarios in part (a) and part (b). If the tension force in part (b) is greater than in part (a), the net force will also be greater, leading to a higher acceleration. This is consistent with Newton's second law:

a = F

_net/m.

Step 5: Conclude that the skier's acceleration in part (b) is greater because the boat applies a larger tension force, which increases the net force acting on the skier. The frictional force remains constant since it depends only on the skier's weight and the coefficient of kinetic friction, which do not change.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This relationship can be expressed with the formula F = ma, where F is the net force, m is the mass, and a is the acceleration. In the context of the skier, understanding how the forces acting on the skier (like tension from the boat and friction) affect acceleration is crucial.

Friction and its Coefficient

Friction is the force that opposes the relative motion of two surfaces in contact. The coefficient of kinetic friction (μₖ) quantifies this force, with higher values indicating greater resistance. In this scenario, the skier experiences kinetic friction with the water, which affects the net force and thus the acceleration. A lower frictional force allows for greater acceleration when the same driving force is applied.

Net Force and Acceleration Comparison

The net force acting on an object is the vector sum of all forces, which determines its acceleration. In comparing two scenarios (part a and part b), if the net force in part b is greater due to reduced friction or increased driving force, the skier will accelerate more. Understanding how changes in forces impact acceleration is key to explaining the differences observed in the two parts of the question.

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Related Practice

Textbook Question

A 68-kg water skier is being accelerated by a ski boat on a flat ('glassy') lake. The coefficient of kinetic friction between the skier's skis and the water surface is μₖ = 0.25 (Fig. 5–59). What is the skier's acceleration if the rope pulling the skier behind the boat applies a horizontal tension force of magnitude F_T = 240N to the skier (θ = 0°)?

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Textbook Question

A 68-kg water skier is being accelerated by a ski boat on a flat ('glassy') lake. The coefficient of kinetic friction between the skier's skis and the water surface is μₖ = 0.25 (Fig. 5–59). What is the skier's horizontal acceleration if the rope pulling the skier exerts a force of F_T = 240N on the skier at an upward angle θ = 12°?

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Textbook Question

A car starts rolling down a 1-in-4 hill (1-in-4 means that for each 4 m traveled along the road, the elevation change is 1 m). How fast is it going when it reaches the bottom after traveling 55 m? Assume an effective coefficient of rolling friction equal to 0.10.

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