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Chapter 3: Problem 377

A car turns a corner on a slippery road at a constant speed of $10 \mathrm{~m} / \mathrm{s}\(. If the coefficient of friction is \)0.5$, the minimum radius of the arc at which the car turns is meter. (A) 20 (B) 10 (C) 5 (D) 4

Short Answer

Expert verified
The minimum radius the car can turn at without sliding is approximately 20.4 meters, so the closest answer among the options is (A) 20.

Step by step solution

01

1. Identify the forces acting on the car

While the car is turning, there are two main forces acting on it: the gravitational force (weight) acting vertically downward, and the friction force acting horizontally to oppose the car's motion. As the car is moving in a curved path, the centripetal force will be provided by the friction force.
02

2. Calculate the centripetal force required

We need to calculate the centripetal force keeping the car in its circular path. The formula for centripetal force is: \[ F_c = m * a_c \] where \(F_c\) is the centripetal force, \(m\) is the mass of the car, and \(a_c\) is the centripetal acceleration. The centripetal acceleration can be calculated using the formula: \[ a_c = \frac{v^2}{r} \] where \(v\) is the speed of the car, which is 10 m/s, and \(r\) is the radius of the arc.
03

3. Calculate the maximum friction force

The maximum friction force available to provide the centripetal force is given by: \[ F_f = \mu * m * g \] where \(F_f\) is the friction force, \(\mu\) is the coefficient of friction (0.5), \(m\) is the mass of the car, and \(g\) is the acceleration due to gravity (approximately \(9.8 \mathrm{m/s^2}\)).
04

4. Set the centripetal force equal to the maximum friction force

Since the centripetal force must be provided by the friction force, we can set the two forces equal to each other: \[ m * a_c = \mu * m * g \] The mass of the car (\(m\)) can be canceled out from both sides, leaving: \[ a_c = \mu * g \] Plug in the given values for \(\mu\) and \(g\) to find \(a_c\): \[ a_c = 0.5 * 9.8 = 4.9 \mathrm{m/s^2} \]
05

5. Solve for the radius of the arc

Now that we have the centripetal acceleration (\(a_c\)), we can find the radius of the arc by plugging it back into the centripetal acceleration formula: \[ 4.9 = \frac{10^2}{r} \] \[ r = \frac{10^2}{4.9} \] \[ r \approx 20.4 \mathrm{m} \] The minimum radius the car can turn at without sliding is approximately 20.4 meters, so the closest answer among the options is: (A) 20

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Most popular questions from this chapter

A wagon weighing \(1000 \mathrm{~kg}\) is moving with a velocity $50 \mathrm{~km} \mathrm{~h}^{-1}\( on smooth horizontal rails. A mass of \)250 \mathrm{~kg}$ is dropped into it. The velocity with which it moves now is (A) \(2.5 \mathrm{~km} \mathrm{~h}^{-1}\) (B) \(20 \mathrm{~km} \mathrm{~h}^{-1}\) (C) \(40 \mathrm{~km} \mathrm{~h}^{-1}\) (D) \(50 \mathrm{~km} \mathrm{~h}^{-1}\)

Newton's third law of motion leads to the law of conservation of (A) Angular momentum (B) Energy (C) mass (D) momentum

A body of mass \(5 \mathrm{~kg}\) starts motion form the origin with an initial velocity $\mathrm{v}_{0} \rightarrow=30 \mathrm{i}+40 \mathrm{j} \mathrm{m} / \mathrm{s}$ If a constant force \(\mathrm{F}=-\left(\mathrm{i}^{\wedge}+5 \mathrm{j}\right) \mathrm{N}\) acts on the body, than the time in which the Y-component of the velocity becomes zero is (A) \(5 \mathrm{~s}\) (B) \(20 \mathrm{~s}\) (C) \(40 \mathrm{~s}\) (D) \(80 \mathrm{~s}\)

Formula for true force is (A) \(\mathrm{F}=\mathrm{ma}\) (B) \(\mathrm{F}=[\\{\mathrm{d}(\mathrm{mv})\\} / \mathrm{dt}]\) (C) \(\mathrm{F}=\mathrm{m}(\mathrm{dv} / \mathrm{dt})\) (D) \(F=m\left(d^{2} x / d t^{2}\right)\)

Three Forces \(F_{1}, F_{2}\), and \(F_{3}\) together keep a body in equilibrium. If \(F_{1}=3 \mathrm{~N}\) along the positive \(\mathrm{X}\) - axis, \(\mathrm{F}_{2}=4 \mathrm{~N}\) along the positive Y-axis then the third force \(F_{3}\) is (A) \(5 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(3 / 4)\) with negative \(\mathrm{y}\) -axis (B) \(5 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(4 / 3)\) with negative \(\mathrm{y}\) -axis (C) \(7 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(3 / 4)\) with negative \(\mathrm{y}\) -axis (D) \(7 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(4 / 3)\) with negative \(\mathrm{y}\) -axis
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