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The elapsed time for a top fuel dragster to start from rest and travel in a straight line a distance of \(\frac{1}{4}\) mile \((402 \mathrm{~m})\) is 4.41 s. Find the minimum coefficient of friction between the tires and the track needed to achieve this result. (Note that the minimum coefficient of friction is found from the simplifying assumption that the dragster accelerates with constant

Short Answer

Expert verified
Answer: The minimum coefficient of friction needed is 4.27.

Step by step solution

01

Calculate the acceleration

Since the dragster starts from rest, the initial velocity (u) is 0. Using the given values of distance (d) and elapsed time (t), we can calculate the acceleration. Use the following kinematic equation: \(v^2 = u^2 + 2ad\) Since the initial velocity (u) is 0, the formula simplifies to: \(v^2 = 2ad\) Now, we can use the given values and rearrange the equation to find the acceleration (a): \(a = \frac{v^2}{2d}\) We do not have the value for final velocity (v) yet. But we can find it by using the equation: \(v = u + at\) Since the initial velocity (u) is 0, the formula simplifies to: \(v = at\) Rearrange the equation to solve for acceleration (a): \(a = \frac{v}{t}\)
02

Calculate the final velocity (v)

To find the final velocity (v), we can use the given values of distance (d) and the elapsed time (t), along with the previously calculated acceleration (a). Use the following kinematic equation: \(v = at\) Plug in the known values: \(v = (\frac{v}{t}) * t\) Rearrange the equation and solve for the final velocity (v): \(v = 402 * \frac{2}{4.41}\) \(v ≈ 182.31 \mathrm{~m/s}\)
03

Calculate the acceleration (a)

Now that we have the final velocity (v), we can calculate the acceleration (a). Use the kinematic equation: \(a = \frac{v^2}{2d}\) Plug in the known values: \(a = \frac{(182.31)^2}{2 * 402}\) \(a ≈ 41.82 \mathrm{~m/s^2}\)
04

Calculate the minimum coefficient of friction (μ)

To find the minimum coefficient of friction (μ), we will use Newton's second law: \(ΣF_{x} = ma\) Since the only horizontal force acting on the dragster is the friction, we have: \(F_{friction} = ma\) Now, use the formula for friction: \(F_{friction} = μ * F_{normal}\) Since the dragster is on a flat surface, the normal force (F_{normal}) is equal to its weight (mg): \(μ * mg = ma\) Rearrange the equation to solve for the coefficient of friction (μ): \(μ = \frac{a}{g}\) Plug in the known values: \(μ = \frac{41.82}{9.81}\) \(μ ≈ 4.27\) The minimum coefficient of friction between the tires and the track needed to achieve this result is 4.27.

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Key Concepts

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

Kinematic Equations
When studying the motion of objects, particularly in physics problems involving constant acceleration, kinematic equations are invaluable tools for connecting various aspects of motion. These equations relate displacement, initial velocity, final velocity, acceleration, and time without the need for knowing the forces involved.

The first key equation used in the solution is the final velocity squared equation: $$v^2 = u^2 + 2ad$$. This relation allows us to calculate the acceleration of the dragster if we know the distance covered and the final velocity - which in turn can be calculated from the elapsed time and the initial velocity through the equation $$v = u + at$$. With the acceleration determined, we can infer further properties of the system, such as the force required to maintain this acceleration, which leads us to discuss concepts like friction and Newton's second law.
Newton's Second Law
Newton's second law is foundational to understanding the dynamics of motion. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration: $$\text{\textbf{F}} = m \times a$$. This law connects the concepts of motion with the forces influencing it and provides a way of quantifying the amount of force required to create a change in an object's motion.

In the context of our dragster problem, after kinematic equations have been used to calculate the acceleration, Newton's second law helps us determine the force of friction. The force of friction, which is the product of the coefficient of friction and the normal force, is the only force moving the dragster forward on a flat surface. The minimum coefficient of friction required between the tires and the track is found by setting the force of friction equal to the product of the mass and acceleration of the dragster. This approach simplifies the problem and allows students to see how force, mass, and acceleration are all part of the same physical reality.
Mechanics of Motion
Mechanics of motion is the branch of physics that delves into the behavior of physical bodies when subjected to forces or displacements and their subsequent effects on the surrounding environment. It is split into kinematics, dealing with motion without considering its causes, and dynamics, which includes the forces that cause motion.

In solving our problem, we applied mechanics of motion by first using kinematic equations to find the speed and acceleration of the dragster. After these quantities were established, we stepped into the realm of dynamics by applying Newton's second law to relate the acceleration to the forces at play, specifically the frictional force needed to support this acceleration. Through these steps, we can appreciate how mechanics, both kinematics, and dynamics, provides a comprehensive framework for analyzing motion and predicting the requisite conditions for a system to behave in a certain way, like determining the minimum coefficient of friction for a dragster to complete a quarter-mile run.

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Most popular questions from this chapter

-4.44 A mass \(m_{1}=20.0 \mathrm{~kg}\) on a frictionless ramp is attached to a light string. The string passes over a frictionless pulley and is attached to a hanging mass \(m_{2}\). The ramp is at an angle of \(\theta=30.0^{\circ}\) above the horizontal. \(m_{1}\) moves up the ramp uniformly (at constant speed). Find the value of \(m_{2}\)

offee filters behave Ince small parachutes, with a drag force that is proportional to the velocity squared, \(F_{\text {drag }}=K v^{2}\). A single coffee filter, when dropped from a height of \(2.0 \mathrm{~m}\), reaches the ground in a time of \(3.0 \mathrm{~s}\). When a second coffee filter is nestled within the first, the drag force remains the same, but the weight is doubled. Find the time for the combined filters to reach the ground. (Neglect the brief period when the filters are accelerating up to their terminal speed.)

\- 4.49 A large cubical block of ice of mass \(M=64 \mathrm{~kg}\) and sides of length \(L=0.40 \mathrm{~m}\) is held stationary on a frictionless ramp. The ramp is at an angle of \(\theta=26^{\circ}\) above the horizontal. The ice cube is held in place by a rope of negligible mass and length \(l=1.6 \mathrm{~m}\). The rope is attached to the surface of the ramp and to the upper edge of the ice cube, a distance \(I\) above the surface of the ramp. Find the tension in the rope.

Two blocks are stacked on a frictionless table, and a horizontal force \(F\) is applied to the top block (block 1). Their masses are \(m_{1}=2.50 \mathrm{~kg}\) and \(m_{2}=3.75 \mathrm{~kg}\). The coefficients of static and kinetic friction between the blocks are 0.456 and 0.380 , respectively a) What is the maximum applied force \(F\) for which \(m_{1}\) will not slide off \(m_{2} ?\) b) What are the accelerations of \(m_{1}\) and \(m_{2}\) when \(F=24.5 \mathrm{~N}\) is applied to \(m_{1}\) ?

If the forces that two interacting objects exert on each other are always exactly equal in magnitude and opposite in direction, how is it possible for an object to accelerate?

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