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 102b

Chapter 5, Problem 102b

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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Determine the angle of the incline using the given slope ratio (1-in-4). The slope ratio indicates that for every 4 m of horizontal distance, the vertical elevation change is 1 m. Use the trigonometric relationship \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \), where \( \tan(\theta) = \frac{1}{4} \). Solve for \( \theta \) using \( \theta = \arctan(\frac{1}{4}) \).

Calculate the gravitational force component along the incline. The force due to gravity along the incline is given by \( F_{\text{gravity, incline}} = m g \sin(\theta) \), where \( m \) is the mass of the car, \( g \) is the acceleration due to gravity (9.8 m/s²), and \( \sin(\theta) \) is the sine of the incline angle.

Account for the force of rolling friction. The rolling friction force is given by \( F_{\text{friction}} = \mu_{r} m g \cos(\theta) \), where \( \mu_{r} \) is the coefficient of rolling friction (0.10), and \( \cos(\theta) \) is the cosine of the incline angle.

Determine the net force acting on the car. The net force is the difference between the gravitational force component along the incline and the rolling friction force: \( F_{\text{net}} = F_{\text{gravity, incline}} - F_{\text{friction}} \).

Use the work-energy principle to find the final velocity. The work done by the net force is equal to the change in kinetic energy. The work done is \( W = F_{\text{net}} \cdot d \), where \( d \) is the distance traveled along the incline (55 m). The change in kinetic energy is \( \Delta KE = \frac{1}{2} m v^2 - 0 \), where \( v \) is the final velocity. Equate \( W \) to \( \Delta KE \) and solve for \( v \): \( v = \sqrt{\frac{2 F_{\text{net}} d}{m}} \). Note that the mass \( m \) cancels out in the equation.

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Key Concepts

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

Inclined Plane Dynamics

The dynamics of an inclined plane involve understanding how gravity acts on an object moving down a slope. The angle of the incline affects the gravitational force component acting parallel to the slope, which influences the object's acceleration. In this case, the 1-in-4 slope indicates a specific angle that can be calculated using trigonometric functions, which is essential for determining the car's acceleration down the hill.

Kinematics

Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. Key equations, such as those relating distance, velocity, and acceleration, are used to analyze the car's motion as it travels down the hill. By applying kinematic equations, one can calculate the final velocity of the car after it has traveled a certain distance.

Friction and Rolling Resistance

Friction, particularly rolling friction, plays a significant role in the motion of objects like cars. The coefficient of rolling friction quantifies the resistance encountered by the car as it rolls down the hill. This frictional force opposes the motion and must be accounted for when calculating the net acceleration and final speed of the car, as it reduces the effective force acting on the car due to gravity.

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Related Practice

Textbook Question

A train traveling at a constant speed rounds a curve of radius 215 m. A lamp suspended from the ceiling swings out to an angle of 18.5° throughout the curve. What is the speed of the train? [Hint: See Example 4–15.]

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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 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

The 70.0-kg climber in Fig. 5–53 is supported in the 'chimney' by the friction forces exerted on his shoes and back. The static coefficients of friction between his shoes and the wall, and between his back and the wall, are 0.80 and 0.60, respectively. What is the minimum normal force he must exert? Assume the walls are vertical and that the static friction forces are both at their maximum. Ignore his grip on the rope.

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Textbook Question

A 28.0-kg block is connected to an empty 2.00-kg bucket by a cord running over a frictionless pulley (Fig. 5–56). The coefficient of static friction between the table and the block is 0.42 and the coefficient of kinetic friction between the table and the block is 0.34. Sand is gradually added to the bucket until the system just begins to move. Calculate the acceleration of the system.

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). Explain why the skier's acceleration in part (b) is greater than that in part (a).

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