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 68

Chapter 7, Problem 68

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?

Verified step by step guidance

Recognize that the stopping distance of a car is determined by the work-energy principle, which states that the work done by the braking force is equal to the car's initial kinetic energy. The formula for kinetic energy is:

E_{k} = (1/2)mv^2

, where m is the mass of the car and v is its velocity.

The work done by the braking force is given by:

W = Fd

, where F is the braking force and d is the stopping distance. Assuming the braking force F is constant, the stopping distance can be expressed as:

d = (1/2)mv^2 / F

Now, consider the case where the velocity is doubled to

2v

. Substitute

v = 2v

into the stopping distance formula:

d' = (1/2)m(2v)^2 / F

Simplify the expression for

d'

d' = (1/2)m(4v^2) / F = 4((1/2)mv^2 / F)

. Notice that this is four times the original stopping distance

d

Conclude that if the car's velocity is doubled, the minimum stopping distance will increase by a factor of 4, assuming the braking force remains constant.

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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 physics that describes the motion of objects without considering the forces that cause the motion. It involves concepts such as velocity, acceleration, and displacement. In this context, understanding how velocity affects stopping distance is crucial, as it relates to the equations of motion that govern an object's behavior when it comes to a stop.

Stopping Distance

Stopping distance is the total distance a vehicle travels from the moment the brakes are applied until it comes to a complete stop. It is influenced by factors such as initial speed, reaction time, and braking force. The relationship between speed and stopping distance is quadratic, meaning that if the speed doubles, the stopping distance increases by a factor of four, which is essential for solving the given problem.

Energy and Work

The concepts of energy and work are fundamental in understanding how a vehicle stops. The kinetic energy of the car, which is proportional to the square of its velocity, must be dissipated through work done by the brakes to bring the car to a stop. This relationship highlights how changes in speed directly affect the energy that needs to be managed, thereby influencing the stopping distance.

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The Work-Energy Theorem

Related Practice

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At room temperature, an oxygen molecule, with mass of 5.31 x 10⁻²⁶ kg, typically has a kinetic energy of about 6.21 x 10⁻²¹ J . How fast is it moving?

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Textbook Question

A 3.5-kg object moving in two dimensions initially has a velocity $v sub 1 right arrow sub$ = (10.0 î + 20.0 ĵ) m/s. A net force $\vec{F}$ then acts on the object for 2.0 s, after which the object’s velocity is $v sub 2 right arrow sub$ = (15.0 î + 30.0 ĵ) m/s. Determine the work done by $\vec{F}$ 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 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 3.0-m-long steel chain is stretched out along the top level of a horizontal scaffold at a construction site, in such a way that 2.0 m of the chain remains on the top level and 1.0 m hangs vertically, Fig. 7–27. At this point, the force on the hanging segment is sufficient to pull the entire chain over the edge. Once the chain is moving, the kinetic friction is so small that it can be neglected. How much work is performed on the chain by the force of gravity as the chain falls from the point where 2.0 m remains on the scaffold to the point where the entire chain has left the scaffold? (Assume that the chain has a linear weight density of 24 N/m.)

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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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