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Ch 09: Work and Kinetic Energy
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 09: Work and Kinetic EnergyProblem 49b
Chapter 9, Problem 49b

A 50 kg ice skater is gliding along the ice, heading due north at 4.0 m/s. The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but μk = 0. Suddenly, a wind from the northeast exerts a force of 4.0 N on the skater. What is the minimum value of μs that allows her to continue moving straight north?

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Step 1: Identify the forces acting on the skater. The skater is moving due north, and a wind from the northeast exerts a force of 4.0 N. The ice provides a static friction force to counteract the sideways component of the wind force. Since μk = 0, the kinetic friction is negligible, and only static friction is relevant.
Step 2: Break the wind force into components. The wind is coming from the northeast, which means it has both an eastward and northward component. Use trigonometry to find the components of the wind force: \( F_{east} = F \cdot \cos(45^\circ) \) and \( F_{north} = F \cdot \sin(45^\circ) \), where \( F = 4.0 \, \text{N} \).
Step 3: Determine the role of static friction. The static friction force must counteract the eastward component of the wind force to prevent the skater from slipping sideways. The maximum static friction force is given by \( F_{friction} = \mu_s \cdot F_{normal} \), where \( F_{normal} \) is the normal force exerted by the ice. For the skater, \( F_{normal} = m \cdot g \), where \( m = 50 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
Step 4: Set up the inequality for static friction. To ensure the skater continues moving straight north, the static friction force must be at least equal to the eastward component of the wind force: \( \mu_s \cdot F_{normal} \geq F_{east} \). Substitute \( F_{normal} = m \cdot g \) and \( F_{east} = F \cdot \cos(45^\circ) \) into the inequality.
Step 5: Solve for \( \mu_s \). Rearrange the inequality to isolate \( \mu_s \): \( \mu_s \geq \frac{F \cdot \cos(45^\circ)}{m \cdot g} \). Substitute the known values for \( F \), \( m \), and \( g \) to find the minimum value of \( \mu_s \).

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Key Concepts

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

Friction

Friction is the force that opposes the relative motion of two surfaces in contact. It is characterized by two coefficients: static friction (μs), which prevents motion, and kinetic friction (μk), which acts when surfaces are sliding against each other. In this scenario, static friction is crucial as it must be sufficient to counteract the lateral force exerted by the wind, allowing the skater to maintain a straight path.
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Force and Motion

Newton's laws of motion describe the relationship between the forces acting on an object and its motion. The first law states that an object at rest or in uniform motion will remain so unless acted upon by a net external force. In this case, the wind exerts a force that could change the skater's direction, necessitating a balance of forces to keep her moving straight north.
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Equilibrium

Equilibrium occurs when the net force acting on an object is zero, resulting in no change in motion. For the skater to continue moving straight north despite the wind's force, the static friction must equal the wind's force. This balance ensures that the skater does not slip sideways, highlighting the importance of calculating the minimum static friction coefficient required for equilibrium.
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Textbook Question

A 50 kg ice skater is gliding along the ice, heading due north at 4.0 m/s. The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but μk = 0. Suddenly, a wind from the northeast exerts a force of 4.0 N on the skater. Use work and energy to find the skater's speed after gliding 100 m in this wind.

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A 737-800 jet airliner has twin engines, each with 105 kN thrust. A 78,000 kg jet reaches a takeoff speed of 70 m/s in a distance of 1100 m. What is the increase in thermal energy due to rolling friction and air drag?