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Ch. 08 - Conservation of Energy
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. 08 - Conservation of EnergyProblem 35
Chapter 8, Problem 35

You slide down an 8.0-m-high icy hill (≈ frictionless). At the bottom is a level stretch where the coefficient of kinetic friction is 0.30. How far would you travel across the level stretch?

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Determine the potential energy at the top of the hill using the formula: Ui = mgh, where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the hill (8.0 m).
At the bottom of the hill, all the potential energy is converted into kinetic energy. Use the conservation of energy principle: Ui = Kf, where Kf = (1/2)mv². Solve for the velocity v at the bottom of the hill.
On the level stretch, the kinetic energy is dissipated by the work done against friction. The work-energy principle states: Kf = W, where W = f_k d. Here, f_k is the kinetic friction force and d is the distance traveled.
Calculate the kinetic friction force using the formula: f_k = \, \(\mu\)_k \(\cdot\) m \(\cdot\) g, where \(\mu\)_k is the coefficient of kinetic friction (0.30).
Combine the equations to solve for the distance d: d = \(\frac{(1/2)mv^2}{\mu_k \cdot m \cdot g}\). Notice that the mass m cancels out, simplifying the calculation. Substitute the known values to find d.

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Key Concepts

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

Potential Energy

Potential energy is the energy stored in an object due to its position in a gravitational field. In this scenario, as you slide down the 8.0-m-high hill, gravitational potential energy is converted into kinetic energy. The initial potential energy can be calculated using the formula PE = mgh, where m is mass, g is the acceleration due to gravity, and h is the height.
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Kinetic Energy and Energy Conservation

Kinetic energy is the energy of an object in motion, given by the formula KE = 0.5mv². As you descend the hill, the potential energy converts into kinetic energy, and at the bottom, all potential energy is transformed into kinetic energy. The principle of conservation of energy states that the total energy in a closed system remains constant, allowing us to equate the potential energy at the top with the kinetic energy at the bottom.
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Friction and Deceleration

Friction is a force that opposes motion between two surfaces in contact. On the level stretch, the coefficient of kinetic friction (0.30) determines the frictional force acting against your motion. This frictional force causes deceleration, which can be calculated using Newton's second law (F = ma), allowing us to determine how far you will slide before coming to a stop.
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Related Practice
Textbook Question

Consider the track shown in Fig. 8–39. The section AB is one quadrant of a circle of radius 2.0 m and is frictionless. B to C is a horizontal span 3.0 m long with a coefficient of kinetic friction μₖ = 0.25. The section CD under the spring is frictionless. A block of mass 1.0 kg is released from rest at A. After sliding on the track, it compresses the spring by 0.20 m. Determine the velocity of the block at point C.

Textbook Question

Consider the track shown in Fig. 8–39. The section AB is one quadrant of a circle of radius 2.0 m and is frictionless. B to C is a horizontal span 3.0 m long with a coefficient of kinetic friction μₖ = 0.25. The section CD under the spring is frictionless. A block of mass 1.0 kg is released from rest at A. After sliding on the track, it compresses the spring by 0.20 m. Determine the velocity of the block at point B.

Textbook Question

A skier of mass m starts from rest at the top of a solid sphere of radius r and slides down its frictionless surface. If friction is present, does the skier fly off at a greater or lesser angle?

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Textbook Question

Chris jumps off a bridge with a 15-m-long bungee cord (a heavy stretchable cord) tied around his ankle, Fig. 8–37. He falls 15 m before the bungee cord begins to stretch. Chris’s mass is 75 kg and we assume the cord obeys Hooke’s law, F = -kx with k = 55 N/m. If we neglect air resistance, estimate what distance d below the bridge Chris’s foot will be before coming to a stop. Ignore the mass of the cord (not realistic, however) and treat Chris as a particle.

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Textbook Question

The 9.0-kg mass in Fig. 8–36 is held just barely in contact with a spring for which k = 450 N/m . When that mass is released, it falls, compressing the spring and pulling the 3.0-kg mass up. How far does the 9.0-kg mass fall before momentarily coming to rest? Ignore friction in the pulley.

Textbook Question

Consider the track shown in Fig. 8–39. The section AB is one quadrant of a circle of radius 2.0 m and is frictionless. B to C is a horizontal span 3.0 m long with a coefficient of kinetic friction μₖ = 0.25. The section CD under the spring is frictionless. A block of mass 1.0 kg is released from rest at A. After sliding on the track, it compresses the spring by 0.20 m. Determine the thermal energy produced as the block slides from B to C.