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Chapter 5: Problem 88

A 12.0-kg box rests on the level bed of a truck. The coefficients of friction between the box and bed are \(\mu_s = 0.19\) and \(\mu_k = 0.15\). The truck stops at a stop sign and then starts to move with an acceleration of 2.20 m/s\(^2\). If the box is 1.80 m from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck? How far does the truck travel in this time?

Short Answer

Expert verified
The box falls off the truck after 2.24 seconds, and the truck travels 5.53 meters.

Step by step solution

01

Determine Maximum Static Frictional Force

Calculate the maximum static friction using the formula \( f_s = \, \mu_s \, \cdot m \, \cdot g \), where \( \mu_s = 0.19 \), \( m = 12.0 kg \), and \( g = 9.81 m/s^2 \). \[ f_s = 0.19 \cdot 12.0 \cdot 9.81 = 22.374 \, N \]
02

Calculate Acceleration Due to Friction

Using Newton's second law, determine the acceleration \( a_s \) that would be caused by the static friction if it were the only force acting: \[ f_s = m \cdot a_s \] \[ 22.374 = 12.0 \cdot a_s \] \[ a_s = \frac{22.374}{12.0} = 1.865 \, m/s^2 \]
03

Compare Forces to Determine If Box Slides

Since the truck has an acceleration of 2.20 m/s two , which is greater than the maximum acceleration due to static friction (1.865 m/s two ), the box will slide. We now need to consider kinetic friction.
04

Calculate Kinetic Friction Force

Calculate the kinetic friction using the formula \( f_k = \mu_k \cdot m \cdot g \), where \( \mu_k = 0.15 \): \[ f_k = 0.15 \cdot 12.0 \cdot 9.81 = 17.658 \, N \]
05

Determine Acceleration of the Box

Using Newton's second law, calculate the actual acceleration \( a_b \) of the box while the truck accelerates: \[ f_k = m \cdot (a_{truck} - a_b) \] \[ 17.658 = 12 \cdot (2.20 - a_b) \] Solving for \( a_b \): \[ a_b = 2.20 - \frac{17.658}{12} \approx 0.72 \, m/s^2 \]
06

Calculate Time to Fall Off Truck

Use the equation for motion \( d = \frac{1}{2} \cdot a_b \cdot t^2 \) where \( d = 1.80 \, m \). Solve for \( t \): \[ 1.80 = \frac{1}{2} \cdot 0.72 \cdot t^2 \] \[ t^2 = \frac{1.80}{0.36} = 5 \] \[ t = \sqrt{5} \approx 2.24 \, s \]
07

Calculate Distance Traveled by Truck

To find how far the truck travels, use the formula: \( d_{truck} = \frac{1}{2} \cdot a_{truck} \cdot t^2 \). \[ d_{truck} = \frac{1}{2} \cdot 2.20 \cdot 2.24^2 \] \[ d_{truck} = 5.53 \, m \]

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Key Concepts

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

Static Friction
Static friction is the force that keeps an object at rest when it is placed on a surface. It acts to prevent the object from moving until a sufficient external force is applied. The maximum static frictional force, which defines the limit before the object starts sliding, is given by the equation:
  • \( f_s = \mu_s \times m \times g \)
Here:
  • \( \mu_s \) is the coefficient of static friction, which depends on the nature of the surfaces in contact.
  • \( m \) is the mass of the object.
  • \( g \) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\) on Earth).
In our example, the maximum static frictional force is calculated to be 22.374 N, which is the threshold that must be overcome for the box to start moving. If the applied force exceeds this amount, even slightly, static friction can no longer hold the box still.
Kinetic Friction
Once the static friction threshold is overcome, an object begins to slide, and kinetic friction takes over. Kinetic friction is generally lower than static friction and acts to oppose the motion of the object as it continues sliding. The force of kinetic friction is calculated by:
  • \( f_k = \mu_k \times m \times g \)
For the box in the problem, kinetic friction is calculated as 17.658 N.
This force opposes the box's movement as it slides along the truck bed.
Kinetic friction is crucial when considering how quickly a moving object comes to a stop or keeps moving when another force is acting on it.
Newton's Second Law
Newton's Second Law of Motion is fundamental in understanding how forces cause motion. It states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
  • \( F = m \times a \)
In our context, this law helps us determine the accelerations involved in moving the box.
Step 2 of the solution uses this law to find the maximum acceleration the static friction can provide (found to be 1.865 m/stwo).
Since the truck's acceleration of 2.20 m/stwoexceeds this, the box begins to slide.
Acceleration
Acceleration refers to how quickly the velocity of an object changes over time. In this problem, both the truck and the box experience different accelerations.
  • The truck's acceleration is 2.20 m/s two , which initiates the box's movement.
  • However, due to the opposing force of kinetic friction, the box experiences a reduced acceleration.
Using kinetic friction and Newton's Second Law, we find that the box accelerates at 0.72 m/s two .
This acceleration determines how quickly the box will slide off the truck, and we use it to calculate the time of 2.24 seconds before the box falls off.
Understanding these different accelerations gives insights into the motion dynamics of the system.

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Most popular questions from this chapter

A stone with mass 0.80 kg is attached to one end of a string 0.90 m long. The string will break if its tension exceeds 60.0 N. The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. (a) Draw a freebody diagram of the stone. (b) Find the maximum speed the stone can attain without the string breaking.

You are called as an expert witness in a trial for a traffic violation. The facts are these: A driver slammed on his brakes and came to a stop with constant acceleration. Measurements of his tires and the skid marks on the pavement indicate that he locked his car's wheels, the car traveled 192 ft before stopping, and the coefficient of kinetic friction between the road and his tires was 0.750. He was charged with speeding in a 45-mi/h zone but pleads innocent. What is your conclusion: guilty or innocent? How fast was he going when he hit his brakes?

You are standing on a bathroom scale in an elevator in a tall building. Your mass is 64 kg. The elevator starts from rest and travels upward with a speed that varies with time according to \(v(t) =\) (3.0 m/s\(^2\))\(t\) \(+\) (0.20 m/s\(^3\))\(t^2\). When \(t =\) 4.0 s, what is the reading on the bathroom scale?

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