/*! 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 25 A car drives around a curve with... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 25

A car drives around a curve with radius \(410 \mathrm{m}\) at a speed of $32 \mathrm{m} / \mathrm{s} .\( The road is banked at \)5.0^{\circ} .$ The mass of the car is \(1400 \mathrm{kg}\). (a) What is the frictional force on the car? (b) At what speed could you drive around this curve so that the force of friction is zero?

Short Answer

Expert verified
Answer: The frictional force acting on the car is 1176.55 N, and the new speed required for the frictional force to be zero is 26.6 m/s.

Step by step solution

01

Determine the gravitational force on the car

First, we need to calculate the gravitational force acting on the car. The gravitational force acting on the car is given by: \(F_g = m \cdot g\), where \(m = 1400\,\mathrm{kg}\) is the mass of the car, and \(g = 9.81\,\mathrm{m/s^2}\) is the acceleration due to gravity. So, \(F_g = 1400\,\mathrm{kg} \cdot 9.81\,\mathrm{m/s^2} = 13734\,\mathrm{N}\).
02

Calculate the centripetal force acting on the car

The centripetal force needed to keep the car moving in a circular path is given by: \(F_c = m\frac{v^2}{r}\), where \(v = 32\,\mathrm{m/s}\) is the speed of the car, and \(r = 410\,\mathrm{m}\) is the radius of the curve. So, \(F_c = 1400\,\mathrm{kg} \cdot \frac{(32\,\mathrm{m/s})^2}{410\,\mathrm{m}} = 3520\,\mathrm{N}\).
03

Determine the normal force acting on the car

The normal force acting on the car is given by \(F_n = F_g \cdot \cos(5^\circ)\). We can now calculate the normal force: \(F_n = 13734\,\mathrm{N} \cdot \cos(5^\circ) = 13645.85\,\mathrm{N}\).
04

Calculate the frictional force on the car for part (a)

By considering the equilibrium of forces acting on the car in the horizontal direction, we can find the frictional force: \(F_f = F_c - F_g \cdot \sin(5^\circ)\) \(F_f = 3520\,\mathrm{N} - 13734\,\mathrm{N} \cdot \sin(5^\circ) = 1176.55\,\mathrm{N}\). So the frictional force on the car is \(1176.55\,\mathrm{N}\).
05

Calculate the new speed required for zero frictional force for part (b)

For the frictional force to be zero, the centripetal force must be equal to the horizontal component of the gravitational force: \(F_c = F_g \cdot \sin(5^\circ)\) We can rewrite this equation in terms of speed: \(v^2 = r \cdot g \cdot \sin(5^\circ)\), Now, we can find the new speed required: \(v = \sqrt{410\,\mathrm{m} \cdot 9.81\,\mathrm{m/s^2} \cdot \sin(5^\circ)} = 26.6\,\mathrm{m/s}\). So, the speed at which the frictional force would be zero is \(26.6\,\mathrm{m/s}\).

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