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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
All textbooksGiancoli Douglas 5th editionCh. 05 - Using Newton's Laws: Friction, Circular Motion, Drag ForcesProblem 104a
Chapter 5, Problem 104a

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 FT = 240N to the skier (θ = 0°)?
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Step 1: Identify the forces acting on the skier. The forces include the tension force (F_T) pulling the skier forward, the kinetic friction force (F_f) opposing the motion, and the gravitational force (F_g) acting vertically downward. The normal force (F_N) acts vertically upward, balancing the gravitational force.
Step 2: Calculate the gravitational force (F_g) acting on the skier. Use the formula: Fg=mg, where m is the mass of the skier (68 kg) and g is the acceleration due to gravity (9.8 m/s²).
Step 3: Determine the kinetic friction force (F_f) using the formula: Ff=μkFN. Since the skier is on a flat surface, the normal force (F_N) equals the gravitational force (F_g). Substitute the values of μk (0.25) and Fg to calculate Ff.
Step 4: Apply Newton's second law of motion in the horizontal direction: Fnet=ma, where Fnet is the net force, m is the mass, and a is the acceleration. The net force is the difference between the tension force (F_T) and the friction force (F_f): Fnet=FT-Ff. Substitute the values of FT (240 N) and Ff to find Fnet.
Step 5: Solve for the acceleration (a) using the formula: a=Fnetm. Substitute the values of Fnet and m (68 kg) to calculate the skier's acceleration.

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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 is expressed by the formula F = ma, where F is the net force, m is the mass, and a is the acceleration. In this scenario, the net force acting on the skier will be the difference between the tension force and the frictional force.
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Frictional Force

Frictional force is the resistance that one surface or object encounters when moving over another. It can be calculated using the formula F_friction = μₖ * N, where μₖ is the coefficient of kinetic friction and N is the normal force. In this case, the normal force is equal to the weight of the skier, which affects the overall acceleration when the skier is being pulled by the boat.
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Net Force

Net force is the total force acting on an object after all the individual forces are combined. It determines the object's acceleration according to Newton's Second Law. In this problem, the net force is the tension force from the boat minus the frictional force opposing the skier's motion, which will ultimately allow us to calculate the skier's acceleration.
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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 horizontal acceleration if the rope pulling the skier exerts a force of FT = 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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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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