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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
All textbooksGiancoli Douglas 5th editionCh. 07 - Work and EnergyProblem 83a
Chapter 7, Problem 83a

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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Step 1: Understand the problem and identify the given values. The mass of the paintball is 33 g (convert to kilograms: 0.033 kg), the maximum speed is 85 m/s, and the barrel length is 32 cm (convert to meters: 0.32 m). The goal is to find the force F using the work-energy principle.
Step 2: Recall the work-energy principle. It states that the work done on an object is equal to the change in its kinetic energy. Mathematically, this is expressed as: W = ΔKE, where W is the work done and ΔKE = KE_{final} - KE_{initial}. Since the paintball starts from rest, KE_{initial} = 0, so W = KE_{final}.
Step 3: Calculate the final kinetic energy of the paintball. The formula for kinetic energy is: KE = \(\frac{1}{2}\)mv^2, where m is the mass and v is the velocity. Substitute the given values for m and v to find KE_{final}.
Step 4: Relate the work done to the force applied. Work is also defined as the product of force and displacement in the direction of the force: W = F \(\cdot\) d, where F is the force and d is the displacement (barrel length). Rearrange this equation to solve for F: F = \(\frac{W}{d}\).
Step 5: Substitute the values for W (from Step 3) and d (barrel length) into the equation for F. This will give the magnitude of the force applied to the paintball by the pressurized gas.

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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 the context of the paintball gun, the work done by the pressurized gas on the paintball as it travels through the barrel converts into kinetic energy, allowing us to calculate the force exerted by the gas. This principle is fundamental in analyzing how energy transfers and transforms in mechanical systems.
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Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion, defined mathematically as KE = 1/2 mv², where m is the mass and v is the velocity of the object. In this scenario, the paintball's kinetic energy at the end of the barrel can be calculated using its maximum speed of 85 m/s. Understanding kinetic energy is crucial for determining how much work is needed to accelerate the paintball to this speed.
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Force and Motion

Force is an interaction that causes an object to change its velocity, described by Newton's second law, F = ma, where F is force, m is mass, and a is acceleration. In the paintball gun, the constant force applied by the pressurized gas accelerates the paintball down the barrel. Analyzing the relationship between force, mass, and acceleration is essential for calculating the force required to achieve the desired speed of the paintball.
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Related Practice
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

A 3.5-kg object moving in two dimensions initially has a velocity v1→\(\overrightarrow{v_1}\)_{} = (10.0 î + 20.0 ĵ) m/s. A net force F→\(\overrightarrow{F}\) then acts on the object for 2.0 s, after which the object’s velocity is v2→\(\overrightarrow{v_2}\)_{} = (15.0 î + 30.0 ĵ) m/s. Determine the work done by F→\(\overrightarrow{F}\) on the object.

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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 car traveling at a velocity v can stop in a minimum distance d. What would be the car’s minimum stopping distance if it were traveling at a velocity of 2v?

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

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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