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

Giancoli Douglas 5th edition

Ch. 07 - Work and Energy

Problem 86a

Chapter 7, Problem 86a

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).

Verified step by step guidance

Convert the mass of the projectile into kilograms: $ m = 25 \ \text{g} = 0.025 \ \text{kg} $. This ensures consistency in SI units.

Determine the acceleration of the projectile as it comes to rest using the kinematic equation: $ v_f^2 = v_i^2 + 2a d $, where $ v_f = 0 \ \text{m/s} $ (final velocity), $ v_i = 360 \ \text{m/s} $ (initial velocity), and $ d = 0.15 \ \text{m} $ (penetration depth). Rearrange to solve for $ a $: $ a = \frac{v_f^2 - v_i^2}{2d} $.

Substitute the known values into the equation to calculate the acceleration $ a $. Note that the acceleration will be negative because the projectile is decelerating.

Use Newton's second law of motion, $ F = ma $, to find the average force exerted by the gel. Here, $ m $ is the mass of the projectile and $ a $ is the acceleration calculated in the previous step.

Substitute the values of $ m $ and $ a $ into the equation $ F = ma $ to compute the average force. Ensure the force is expressed in newtons (N).

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Key Concepts

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

Kinematics

Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. It involves concepts such as displacement, velocity, and acceleration. In this scenario, kinematics helps us understand how the projectile moves through the gel and how its velocity changes as it penetrates.

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, expressed as F = ma. This principle is crucial for calculating the average force exerted by the gel on the projectile as it decelerates to a stop, allowing us to relate the change in momentum to the force experienced.

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 case, as the projectile penetrates the gel, the work done by the gel's resistance force results in a decrease in the projectile's kinetic energy, which can be used to calculate the average force exerted by the gel over the distance of penetration.

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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 $\vec{F}$ is applied in the horizontal direction (so $\vec{F}$ = Fî ), moving the bob very slowly so the acceleration is essentially zero. (Note that the magnitude of $\vec{F}$ 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, $\vec{F}$ , to move the pendulum from θ = 0 to θ₀.

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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 M_S 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)M_S 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( M_S/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

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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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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Textbook Question

The force required to compress an “imperfect” horizontal spring (doesn’t follow Hooke’s law) an amount x is given by F = 150x + 12x³, where x is in meters and F in newtons. If the spring is compressed 2.0 m, what speed will it give to a 3.0-kg ball held against it and then released?

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