/*! 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 32 A pickup truck is carrying a too... [FREE SOLUTION] | 91Ó°ÊÓ

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A pickup truck is carrying a toolbox, but the rear gate of the truck is missing. The toolbox will slide out if it is set moving. The coefficients of kinetic friction and static friction between the box and the level bed of the truck are 0.355 and 0.650 , respectively. Starting from rest, what is the shortest time this truck could accelerate uniformly to \(30.0 \mathrm{~m} / \mathrm{s}\) without causing the box to slide? Draw a free-body diagram of the toolbox.

Short Answer

Expert verified
The shortest time the truck could uniformly accelerate to 30.0 m/s without causing the box to slide can be calculated with the given coefficients of kinetic friction and static friction, and the final velocity. By firstly identifying these and understanding their role in the problem, and then applying Newton's second law, one can find the maximum static friction and thus the maximum acceleration the truck can have whilst still keeping the toolbox from sliding. Finally, the minimum time it would take for the truck to reach the given speed can be calculated using the kinematic equation for final velocity.

Step by step solution

01

Identify knowns and unknowns

Knowns: Final velocity (\(v_f\)) = 30 m/s, Initial velocity (\(v_i\)) = 0 (starting from rest), Coefficients of kinetic friction (\(μ_k\)) = 0.355 and static friction (\(μ_s\)) = 0.650, Acceleration (\(a\)) is uniform but unknown, Time (\(t\)) is unknown.
02

Apply Newton's Second Law

The maximum force of static friction (\(F_{s-max}\)) can be calculated as \(F_{s-max} = μ_s . mg\). This is the maximum force that can be applied to the box without it starting to move. Therefore, the truck's acceleration (\(a\)) must create an inertia force equal to or smaller than \(F_{s-max}\). So, Newton's Second Law gives \(ma = μ_s . mg\). From this equation, we can get the maximum acceleration (\(a_{max}\)): \(a_{max} = μ_s . g\).
03

Find the shortest time

The vehicle must maintain an acceleration equal or smaller than the \(a_{max}\) to prevent the toolbox from sliding. Therefore, the shortest time (\(t_{min}\)) the truck can uniformly accelerate to 30 m/s can be found from the kinematic equation \(v_f = v_i + a . t\). Given that the initial speed is 0, the equation simplifies to \(t_{min} = v_f / a_{max}\).

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Key Concepts

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

Newton's Second Law
When we delve into any physics problem related to motion and forces, Newton's Second Law often serves as our go-to principle. This law states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. This can be elegantly expressed with the equation:

\[ F = ma \]
where F represents the net force applied to the object, m is the object's mass, and a is its acceleration. In the case of the pickup truck and the toolbox, the law is applied to figure out how the maximum static frictional force can determine the highest possible acceleration of the truck without making the toolbox slide off.

Understanding the relationship between force and acceleration allows us to predict and manipulate motion, granting us a powerful tool in solving many real-world physics problems. The key takeaway here is that more force leads to more acceleration, but also that this is only true as long as the motion remains unopposed—in our example, by static friction.
Static Friction
In our everyday lives, we are constantly experiencing and relying on static friction. It's what allows us to walk without slipping, and it keeps objects in place when we don't want them to move.

Static friction is a force that resists the initial movement of two surfaces that are in contact with each other. When we attempt to slide one object over another, like our toolbox on the truck bed, static friction holds it steady up to a certain point. This point is determined by the coefficient of static friction (\mu_s) and the normal force, which in most cases is simply the weight of the object. Mathematically, the maximum static friction force can be described as:

\[ F_{s-max} = °À³Ù±ð³æ³Ùµ÷μ³å²õ°¨ . mg \]
Here, mg signifies the weight of the object. The coefficient °À³Ù±ð³æ³Ùµ÷μ³å²õ°¨ is a dimensionless value unique to the pairing of materials at contact. In our truck problem, the value of °À³Ù±ð³æ³Ùµ÷μ³å²õ°¨ is particularly important; it represents the threshold of force that the toolbox can withstand before sliding commences. Once this threshold is exceeded, static friction can no longer do its job, and the object begins to move, transitioning into kinetic friction.
Kinematic Equations
Much of classical mechanics is concerned with the motion of objects—how they start, stop, and change direction. Kinematic equations allow us to describe this motion quantitatively by relating velocity, acceleration, and time.

One fundamental equation is used frequently for objects with constant acceleration:

\[ v_f = v_i + at \]
Here, v_f represents the final velocity, v_i is the initial velocity, a is the acceleration, and t is the time elapsed. For our truck and toolbox scenario, we rearrange this equation to solve for the time it takes for the truck to reach a certain speed without the toolbox sliding off.

Using these kinematic equations, we can plug in the values we have—like the acceleration, which we deduced using Newton's Second Law and the coefficient of static friction—to find out how quickly the truck can safely reach the speed of 30 m/s. The simplicity yet robustness of kinematic equations makes them invaluable for solving a wide range of problems in physics. Always remember, these equations are best used when acceleration is constant, which is assumed in our truck problem.

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Most popular questions from this chapter

\(\mathrm{A} 50.0 \mathrm{~kg}\) stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane. (a) If the plane's speed at the lowest point of the circle is \(95.0 \mathrm{~m} / \mathrm{s},\) what is the minimum radius of the circle so that the acceleration at this point will not exceed \(4.00 g ?\) (b) What is the apparent weight of the pilot at the lowest point of the pullout?

Some sliding rocks approach the base of a hill with a speed of \(12 \mathrm{~m} / \mathrm{s} .\) The hill rises at \(36^{\circ}\) above the horizontal and has coefficients of kinetic friction and static friction of 0.45 and \(0.65,\) respectively, with these rocks. (a) Find the acceleration of the rocks as they slide up the hill. (b) Once a rock reaches its highest point, will it stay there or slide down the hill? If it stays, show why. If it slides, find its acceleration on the way down.

A light rope is attached to a block with mass \(4.00 \mathrm{~kg}\) that rests on a friction less, horizontal surface. The horizontal rope passes over a friction less, massless pulley, and a block with mass \(m\) is suspended from the other end. When the blocks are released, the tension in the rope is \(15.0 \mathrm{~N}\). (a) Draw two free-body diagrams: one for each block. (b) What is the acceleration of either block? (c) Find \(m\). (d) How does the tension compare to the weight of the hanging block?

Genesis Crash. On September \(8,2004,\) the Genesis spacecraft crashed in the Utah desert because its parachute did not open. The \(210 \mathrm{~kg}\) capsule hit the ground at \(311 \mathrm{~km} / \mathrm{h}\) and penetrated the soil to a depth of \(81.0 \mathrm{~cm}\). (a) What was its acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(g\) 's) assumed to be constant, during the crash? (b) What force did the ground exert on the capsule during the crash? Express the force in newtons and as a multiple of the capsule's weight. (c) How long did this force last?

Two blocks, with masses \(4.00 \mathrm{~kg}\) and \(8.00 \mathrm{~kg},\) are connected by a string and slide down a \(30.0^{\circ}\) inclined plane (Fig. \(\mathbf{P 5 . 9 6}\) ). The coefficient of kinetic friction between the \(4.00 \mathrm{~kg}\) block and the plane is \(0.25 ;\) that between the \(8.00 \mathrm{~kg}\) block and the plane is \(0.35 .\) Calculate (a) the acceleration of each block and (b) the tension in the string. (c) What happens if the positions of the blocks are reversed, so that the \(4.00 \mathrm{~kg}\) block is uphill from the \(8.00 \mathrm{~kg}\) block?

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