/*! 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 62 The minimum velocity (in \(\math... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 62

The minimum velocity (in \(\mathrm{ms}^{-1}\) ) with which a car driver must traverse a flat curve of radius \(150 \mathrm{~m}\) and coefficient of friction \(0.6\) to avoid skidding is (a) \(60 \mathrm{~ms}^{-1}\) (b) \(30 \mathrm{~ms}^{-1}\) (c) \(15 \mathrm{~ms}^{-1}\) (d) \(25 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The minimum velocity is 30 ms鈦宦� (option b).

Step by step solution

01

Identify the formula

The minimum velocity required to traverse a curve without skidding is determined by the formula: \(v = \sqrt{\mu g r}\), where \(\mu\) is the coefficient of friction, \(g\) is the acceleration due to gravity (approximately \(9.8 \, \mathrm{ms}^{-2}\)), and \(r\) is the radius of the curve.
02

Substitute given values

Substitute the given values into the formula: \(\mu = 0.6\), \(g = 9.8 \, \mathrm{ms}^{-2}\), \(r = 150 \, \mathrm{m}\). The equation becomes \(v = \sqrt{0.6 \times 9.8 \times 150}\).
03

Calculate under the square root

Calculate the product under the square root: \(0.6 \times 9.8 \times 150 = 882\).
04

Find the square root

Find the square root of the value calculated in the previous step: \(v = \sqrt{882} \approx 29.7 \, \mathrm{ms}^{-1}\).
05

Round to nearest value

Round the calculated velocity to the nearest value from the options given. The nearest option is \(30 \, \mathrm{ms}^{-1}\), which matches option (b).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding Friction Coefficient
The friction coefficient is a key player in preventing a car from skidding on a curve. Imagine friction as the invisible grip between the tires of your car and the road surface. This grip, quantified by the coefficient of friction, is crucial for maintaining control. A higher friction coefficient means better grip and, consequently, a higher resistance to skidding. In our scenario, the given coefficient of friction is 0.6. This value plays a critical role in determining the minimum velocity at which a car won't skid when navigating a curve.
Basics of Circular Motion
Circular motion is everywhere, from cars taking turns to planets orbiting the sun. When a car travels along a curve, it experiences circular motion, moving along the circumference of a circle with the curve as part of its path. The radius of this circle (150 meters in our problem) is vital because it helps define the path's curvature. In physics, circular motion introduces unique challenges, like maintaining the correct speed to avoid skidding. If the speed is too high, the car can't handle the curve's curvature and might skid outwards.
Skidding Prevention Techniques
Skidding occurs when a vehicle loses traction on the road, leading to a loss of control. Preventing skidding, especially on curves, is vital for safety. The main tools for preventing skidding are controlling speed and ensuring adequate friction. Practical techniques include:
  • Adjusting speed based on curve radius and surface conditions.
  • Ensuring tires are in good condition with appropriate tread.
  • Avoiding sudden maneuvers that might disrupt the vehicle's balance.
Using these strategies, drivers can maintain control and prevent accidents on curves.
Role of Centripetal Force
At the heart of circular motion and skidding prevention is the concept of centripetal force. This force is necessary to pull a vehicle towards the center of its circular path. Without sufficient centripetal force, the vehicle cannot complete the turn and might slide sideways. In our exercise, the centripetal force required to keep the car on its path is provided by friction. Thus, the calculation of the minimum velocity involves ensuring that this force is adequate to match the needed centripetal force, factoring in the friction coefficient and gravitational force.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A block is gently placed on a conveyor belt moving horizontally with constant speed. After 4 s the velocity of the block becomes equal to the velocity of belt. If the coefficient of friction between the block and the belt is 0.2, then velocity of the conveyor belt is (a) \(2 \mathrm{~ms}^{-1}\) (b) \(4 \mathrm{~ms}^{-1}\) (c) \(6 \mathrm{~ms}^{-1}\) (d) \(8 \mathrm{~ms}^{-1}\)

A maximum speed that can be achieved without skidding by a car on a circular unbanked road of radius \(R\) and coefficient of static friction \(\mu\) is (a) \(\mu \mathrm{Rg}\) (b) \(R g \sqrt{\mu}\) (c) \(\mu \sqrt{R g}\) (d) \(\sqrt{\mu \mathbb{R g}}\)

Two masses of \(3 \mathrm{~kg}\) and \(5 \mathrm{~kg}\) are suspended from the ends of an unstreatchable massless cord passing over a frictionless pulley. When the masses are released, the pressure on the pulley is (a) \(2 \mathrm{~kg} \mathrm{f}\) (b) \(7.5 \mathrm{~kg}\) f (c) \(8 \mathrm{~kg} \mathrm{f}\) (d) \(15 \mathrm{kgf}\)

A player caught a cricket ball of mass \(150 \mathrm{~g}\) moving at a rate of \(20 \mathrm{~ms}^{-1}\), If the catching process is completed in \(0.1 \mathrm{~s}\), the force of blow exerted by the ball on the hands of the player is equal to [AIFEE 2006] (a) \(3 \bar{N}\) (b) \(-30 \mathrm{~N}\) (c) \(300 \mathrm{~N}\) (d) \(150 \mathrm{~N}\)

A plate of mass \(M\) is placed on a horizontal frictionless surface (see figure) and a body of mass \(m\) is placed on this plate. The coefficient of dynamics friction between this body and the plate is \(\mu\). If a force \(2 \mu m g\) is applied to the body of mass \(m\) along the horizontal, the acceleration of the plate will be (a) \(\frac{\mu m}{M} g\) (b) \(\frac{\mu m}{(M+m)} g\) (c) \(\frac{2 \mu m}{M} g\) (d) \(\frac{2 \mu m}{(M+m)} g\)
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Magnetism

Read Explanation

Atoms and Radioactivity

Read Explanation

Dynamics

Read Explanation

Classical Mechanics

Read Explanation

Work Energy and Power

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.