/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 The greatest force a level road ... [FREE SOLUTION] | 91Ó°ÊÓ

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The greatest force a level road can exert on the tires of a certain 2000 -kg car is \(4 \mathrm{kN}\). What is the highest speed the car can round a curve of radius \(200 \mathrm{m}\) without skidding?

Short Answer

Expert verified
20 m/s

Step by step solution

01

Identify the Given Variables

The mass of the car, which is given as 2000 kg, and the maximum force exerted by the road on the tires, which is given as 4000 N (since 4 kN = 4000 N). The radius of the curve is 200 m.
02

Understand the Relationship

Realize that centripetal force is required to keep the car moving in a circle. The maximum centripetal force that can be applied without skidding is equal to the limit provided by the road's force on the tires, which is 4000 N.
03

Use the Centripetal Force Formula

Use the formula for centripetal force: \[ F_c = \frac{mv^2}{r} \] where \( F_c \) is the centripetal force, \( m \) is the mass of the car, \( v \) is the velocity, and \( r \) is the radius of the curve.
04

Substitute Known Values

Substitute the known values into the equation: \[ 4000 = \frac{2000 \cdot v^2}{200} \]
05

Solve for Velocity

Solve for \( v \) by isolating it in the equation: \[ 4000 = \frac{2000v^2}{200} \quad \rightarrow \quad 4000 = 10v^2 \quad \rightarrow \quad v^2 = 400 \quad \rightarrow \quad v = \sqrt{400} \quad \rightarrow \quad v = 20 \, m/s \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

circular motion
Circular motion describes the motion of an object traveling along a circular path. One crucial aspect of circular motion is that an object moving in a circle constantly changes its direction of motion. This change in direction requires a continuous inward force directed towards the center of the circular path. This inward force is called centripetal force. Without this force, the object would travel in a straight line due to inertia.
maximum velocity
The maximum velocity of an object in circular motion is the highest speed it can maintain while remaining on its curved path without skidding or slipping. This speed depends on several factors, including the mass of the object, the radius of the curve, and the maximum centripetal force that can be exerted. In the given problem, the maximum force exerted by the road on the car's tires is 4000 N, which limits the car's maximum velocity. By using the centripetal force formula, you can solve for the maximum velocity:
force of friction
The force of friction is the resisting force acting between two surfaces in contact. In the context of our problem, it is the force exerted by the road on the car's tires to prevent skidding. This frictional force provides the necessary centripetal force to keep the car moving along the curve. When the force of friction reaches its maximum value (4000 N in this case), the car is at its limit of adhesion. If the car's speed exceeds this threshold, the frictional force will be insufficient to keep the car on the curved path, causing it to skid.
Newton's laws
Newton's laws of motion form the foundation for understanding motion and forces. These laws are critical in solving circular motion problems like the one given. Newton's First Law, also called the Law of Inertia, states that an object will remain at rest or move in a straight line at a constant speed unless acted upon by an external force. In circular motion, this external force is the centripetal force. Newton's Second Law states that the force exerted on an object is equal to its mass times its acceleration force (F) = mass (m) x acceleration (a)). In circular motion, this acceleration is the centripetal acceleration directed towards the center of the circular path. Newton's Third Law states that for every action, there is an equal and opposite reaction. Thus, the frictional force the road exerts on the tires is matched by the force the tires exert on the road.

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Most popular questions from this chapter

In accelerating from a standing start to a speed of \(300 \mathrm{km} / \mathrm{h}(186 \mathrm{mi} / \mathrm{h}-\text { not its top speed! }),\) the 1900 -kg Bugatti Veyron sports car exerts an average force on the road of \(9.4 \mathrm{kN}\). How long does the car take to reach \(300 \mathrm{km} / \mathrm{h} ?\)

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