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Ch. 07 - Work and Energy
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 7, Problem 87a

An airplane pilot fell 370 m after jumping from an aircraft without his parachute opening. He landed in a snowbank, creating a crater 1.1 m deep, but survived with only minor injuries. Assuming the pilot’s mass was 82 kg and his terminal velocity was 45 m/s, estimate the work done by the snow in bringing him to rest.

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Step 1: Understand the problem. The work done by the snow in bringing the pilot to rest can be calculated using the work-energy principle. The work done by the snow is equal to the change in the pilot's kinetic energy as he comes to rest.
Step 2: Write the expression for the work-energy principle. The work done (W) is given by: W = ΔK, where ΔK is the change in kinetic energy. Since the pilot comes to rest, his final kinetic energy is zero, and the initial kinetic energy is given by: Ki = rac{1}{2}mv2.
Step 3: Substitute the given values into the kinetic energy formula. The pilot's mass is m = 82 kg, and his terminal velocity is v = 45 m/s. Calculate the initial kinetic energy using: Ki = rac{1}{2}mv2.
Step 4: Since the final kinetic energy is zero, the work done by the snow is equal to the negative of the initial kinetic energy: W = -Ki. This negative sign indicates that the snow is doing work to stop the pilot.
Step 5: Conclude that the work done by the snow can now be calculated by substituting the values into the formula. Ensure the units are consistent, and the result will be in joules (J).

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Key Concepts

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

Work-Energy Principle

The Work-Energy Principle states that the work done on an object is equal to the change in its kinetic energy. In this scenario, the work done by the snow on the pilot can be calculated by determining the difference between his kinetic energy just before impact and his kinetic energy after coming to rest. This principle is fundamental in analyzing how forces affect the motion and energy of objects.
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Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, calculated using the formula KE = 0.5 * m * v², where m is the mass and v is the velocity. For the pilot, his kinetic energy just before hitting the snowbank can be determined using his mass and terminal velocity. Understanding kinetic energy is crucial for evaluating the impact forces involved in the landing.
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Terminal Velocity

Terminal velocity is the constant speed an object reaches when the force of gravity is balanced by the drag force acting against it, resulting in no net acceleration. For the pilot, reaching a terminal velocity of 45 m/s means he fell at this speed before impact, which is essential for calculating his kinetic energy and the subsequent work done by the snow to bring him to rest.
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Related Practice
Textbook Question

A simple pendulum consists of a small object of mass m (the “bob”) suspended by a cord of length ℓ (Fig. 7–34) of negligible mass. A force F\(\overrightarrow{F}\) is applied in the horizontal direction (so F\(\overrightarrow{F}\) = Fî ), moving the bob very slowly so the acceleration is essentially zero. (Note that the magnitude of F\(\overrightarrow{F}\) will need to vary with the angle θ that the cord makes with the vertical at any moment.) Determine the work done by this force, F\(\overrightarrow{F}\), to move the pendulum from θ = 0 to θ₀.

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Textbook Question

A 25-g projectile is fired into a cube of ballistic gel at a velocity of 360 m/s. If the projectile penetrates 15 cm into the gel before stopping, find the average force exerted by the gel onto the projectile. Use kinematics and dynamics (Newton's laws).

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Textbook Question

We usually neglect the mass of a spring if it is small compared to the mass attached to it. But in some applications, the mass of the spring must be taken into account. Consider a spring of unstretched length ℓ and mass MS uniformly distributed along the length of the spring. A mass m is attached to the end of the spring. One end of the spring is fixed and the mass m is allowed to vibrate horizontally without friction (Fig. 7–31). Each point on the spring moves with a velocity proportional to the distance from that point to the fixed end. For example, if the mass on the end moves with speed v₀, the midpoint of the spring moves with speed v₀ / 2. Show that the kinetic energy of the mass plus spring when the mass m is moving with velocity v is K = (1/2)Mv² where M = m + (1/3)MS is the “effective mass” of the system. [Hint: Let D be the total length of the stretched spring. Then the velocity of an infinitesimal length dx of spring, of mass dM, located at x is v(x) = v₀(x/D). Note also that dM = dx( MS/D).]

Textbook Question

A package of mass m is placed onto a horizontal conveyor belt moving at speed v (Fig. 7–32). The coefficient of kinetic friction between package and belt is μₖ. What is the package's displacement d during this time?

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Textbook Question

In the game of paintball, players use guns powered by pressurized gas to propel 33-g gel capsules filled with paint at the opposing team. Game rules dictate that a paintball cannot leave the barrel of a gun with a speed greater than 85 m/s. Model the shot by assuming the pressurized gas applies a constant force F to a 33-g capsule over the length of the 32-cm barrel. Determine F by using the work-energy principle.

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