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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

BIO Stay Awake! An astronaut is inside a \(2.25 \times 10^{6} \mathrm{~kg}\) rocket that is blasting off vertically from the launch pad. You want this rocket to reach the speed of sound \((331 \mathrm{~m} / \mathrm{s})\) as quickly as possible, but astronauts are in danger of blacking out at an acceleration greater than \(4 g\). (a) What is the maximum initial thrust this rocket's engines can have but just barely avoid blackout? Start with a free-body diagram of the rocket. (b) What force, in terms of the astronaut's weight \(w\), does the rocket exert on her? Start with a free-body diagram of the astronaut. (c) What is the shortest time it can take the rocket to reach the speed of sound?

A box with mass \(m\) is dragged across a level floor with coefficient of kinetic friction \(\mu_{\mathrm{k}}\) by a rope that is pulled upward at an angle \(\theta\) above the horizontal with a force of magnitude \(F\). (a) In terms of \(m, \mu_{\mathrm{k}}, \theta,\) and \(g,\) obtain an expression for the magnitude of the force required to move the box with constant speed. (b) Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a \(90 \mathrm{~kg}\) patient across a floor at constant speed by pulling on him at an angle of \(25^{\circ}\) above the horizontal. By dragging weights wrapped in an old pair of pants down the hall with a spring balance, you find that \(\mu_{\mathrm{k}}=0.35 .\) Use the result of part (a) to answer the instructor's question.

Two crates connected by a rope lie on a horizontal surface (Fig. E5.37). Crate \(A\) has mass \(m_{A}\), and crate \(B\) has mass \(m_{B}\). The coefficient of kinetic friction between each crate and the surface is \(\mu_{\mathrm{k}} .\) The crates are pulled to the right at constant velocity by a horizontal force \(\overrightarrow{\boldsymbol{F}}\). Draw one or more free-body diagrams to calculate the following in terms of \(m_{A}, m_{B},\) and \(\mu_{\mathrm{k}}:\) (a) the magnitude of \(\overrightarrow{\boldsymbol{F}}\) and \((\mathrm{b})\) the tension in the rope connecting the blocks.

An \(8.00 \mathrm{~kg}\) box sits on a ramp that is inclined at \(33.0^{\circ}\) above the horizontal. The coefficient of kinetic friction between the box and the surface of the ramp is \(\mu_{\mathrm{k}}=0.300 .\) A constant horizontal force \(F=26.0 \mathrm{~N}\) is applied to the box (Fig. \(\mathbf{P} 5.73),\) and the box moves down the ramp. If the box is initially at rest, what is its speed \(2.00 \mathrm{~s}\) after the force is applied?

A steel ball with mass \(m\) is suspended from the ceiling at the bottom end of a light, 15.0 -m-long rope. The ball swings back and forth like a pendulum. When the ball is at its lowest point and the rope is vertical, the tension in the rope is three times the weight of the ball, so \(T=3 m g .\) (a) What is the speed of the ball as it swings through this point? (b) What is the speed of the ball if \(T=m g\) at this point, where the rope is vertical?

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