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Chapter 5: Problem 88

A 12.0 -kg box rests on the flat floor of a truck. The coefficients of friction between the box and floor are \(\mu_{\mathrm{s}}=0.19\) and \(\mu_{\mathrm{k}}=0.15 .\) The truck stops at a stop sign and then starts to move with an acceleration of 2.20 \(\mathrm{m} / \mathrm{s}^{2} .\) If the box is 1.80 \(\mathrm{m}\) 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 after 2.22 seconds, and the truck travels 5.41 meters.

Step by step solution

01

Determine the Maximum Static Friction Force

First, let's calculate the maximum static friction force that can act on the box before it begins to slide. This force is given by the formula \( f_{s\_{max}} = \mu_s \times m \times g \), where \( m = 12.0\, \text{kg} \) and \( g = 9.8\, \text{m/s}^2 \) (acceleration due to gravity).\Substituting the values, we have: \[ f_{s\_{max}} = 0.19 \times 12.0 \times 9.8 = 22.344 \, \text{N}. \]
02

Calculate the Force Needed to Accelerate the Box with the Truck

The force required to accelerate the box with the truck is calculated by \( F = m \times a \), where \( a = 2.20\, \text{m/s}^2 \).\Plugging in the numbers, we find:\[ F = 12.0 \times 2.20 = 26.4 \, \text{N}. \]
03

Check if Static Friction Can Prevent Sliding

Now, compare the maximum static friction force with the force required to accelerate the box. Since \( f_{s\_{max}} = 22.344 \, \text{N} \) is less than \( F = 26.4 \, \text{N} \), the static friction is not sufficient to keep the box moving with the truck without sliding. The box will start to slide.
04

Calculate the Acceleration of the Box

Once the box starts sliding, kinetic friction takes over. Calculate the force of kinetic friction using \( f_k = \mu_k \times m \times g \), where \( \mu_k = 0.15 \).\\( f_k = 0.15 \times 12.0 \times 9.8 = 17.64 \, \text{N}. \)Using Newton's second law, the net force is \( m \times a_{box} = \text{applied force} - f_k \), \( 12.0 \times a_{box} = 26.4 - 17.64 \), \( a_{box} = \frac{26.4 - 17.64}{12.0} = 0.73 \, \text{m/s}^2. \)
05

Calculate the Time to Slide off the Truck

Use the equation \( x = \frac{1}{2} a_{box} t^2 \) where \( x = 1.80 \, \text{m} \) is the distance to slide off. Solve for \( t \):\[ 1.80 = \frac{1}{2} \times 0.73 \times t^2 \]\[ t^2 = \frac{1.80 \times 2}{0.73} = 4.93 \]\[ t = \sqrt{4.93} = 2.22 \, \text{seconds}. \]
06

Calculate the Distance the Truck Travels

The truck moves with an acceleration of 2.20 \( \text{m/s}^2 \). Calculate the distance using \( d = \frac{1}{2} a \times t^2 \): \[ d = \frac{1}{2} \times 2.20 \times (2.22)^2 \]\[ d = 5.41 \, \text{meters}. \]

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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 resistant force that prevents two surfaces from sliding past each other. It's crucial when objects are at rest. The maximum static friction force can be calculated using the equation \( f_{s\_{max}} = \mu_s \times m \times g \), in which \( \mu_s \) is the coefficient of static friction, \( m \) is the mass, and \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)).
It's important to note that static friction isn't a constant value; it can vary up to its maximum limit. If the required force to move an object exceeds this maximum value, the object will start sliding. In our case, the maximum static friction force was found to be \( 22.344 \, \text{N} \). This force was insufficient to hold the box on a moving truck because it required \( 26.4 \, \text{N} \) of force. Thus, the static friction could not prevent the box from sliding.
  • Key Points:
  • Static friction keeps objects stationary.
  • Maximum static friction depends on the normal force and the coefficient of static friction.
  • If the applied force exceeds the maximum static friction, sliding begins.
Kinetic Friction
When an object starts to slide, the force of friction shifts from static to kinetic friction. Kinetic friction is generally less than static friction for the same materials. The formula used is \( f_k = \mu_k \times m \times g \), where \( \mu_k \) is the coefficient of kinetic friction. In our scenario, as the box began to slide, the kinetic friction force was calculated to be \( 17.64 \, \text{N} \).
Understanding the role of kinetic friction is essential in dynamics and physics problems, especially in solving questions about motion on surfaces. Since kinetic friction acts in the opposite direction to movement, it reduces the net force that drives the object. This force needs to be taken into account to correctly calculate an object's acceleration once it is in motion.
  • Key Points:
  • Kinetic friction applies when an object is in motion.
  • The force it provides is lower than static friction for the same surfaces.
  • Essential for calculating net forces and resulting accelerations in motion problems.
Newton's Second Law
Newton's Second Law of Motion is a fundamental principle that describes how motion changes under the influence of forces. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The law is usually expressed by the formula \( F = m \times a \), where \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration.
In our physics problem involving the box and the truck, Newton's second law was used to determine both the force required to accelerate the box and the actual acceleration of the box as it slides. When subtracting the kinetic friction from the force applied by the truck, we found the net force, which was then used to calculate the box's acceleration as \( 0.73 \, \text{m/s}^2 \).
  • Key Points:
  • Acceleration depends on net force and mass.
  • Helps in predicting motion changes due to different forces.
  • Critical for solving dynamics problems in physics.

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Most popular questions from this chapter

A 550 -N physics student stands on a bathroom scale in an 850 -kg (including the student) elevator that is supported by a cable. As the elevator starts moving, the scale reads 450 \(\mathrm{N}\) . (a) Find the acceleration of the elevator (magnitude and direction). (b) What is the acceleration if the scale reads 670 \(\mathrm{N}\) ? (c) If the scale reads zero, should the student worry? Explain. (d) What is the tension in the cable in parts (a) and (c)?

One December identical twins Jena and Jackie are playing on a large merry-go- round (a disk mounted parallel to the ground, on a vertical axle through its center) in their school playground in northern Minnesota. Each twin has mass 30.0 \(\mathrm{kg}\) . The icy coating on the merry-go-round surface makes it frictionless. The merry-go-round revolves at a constant rate as the twins ride on it. Jena, sitting 1.80 \(\mathrm{m}\) from the center of the merry-go-round, must hold on to one of the metal posts attached to the merry-go-round with a horizontal force of 60.0 \(\mathrm{N}\) to keep from sliding off. Jackie is sitting at the edge, 3.60 \(\mathrm{m}\) from the center. (a) With what horizontal force must Jackie hold on to keep from falling off? (b) If Jackie falls off, what will be her horizontal velocity when she becomes airborne?

A 1125 -kg car and a 2250 -kg pickup truck approach a curve on the expressway that has a radius of 225 \(\mathrm{m}\) . (a) At what angle should the highway engineer bank this curve so that vehicles traveling at 65.0 \(\mathrm{mi} / \mathrm{h}\) can safely round it regardless of the condition of their tires? Should the heavy truck go slower than the lighter car? (b) As the car and truck round the curve at find the normal force on each one due to the highway surface.

A box with weight \(w\) is pulled at constant speed along a level floor by a force \(\vec{\boldsymbol{F}}\) that is at an angle \(\theta\) above the horizontal. The coefficient of kinetic friction between the floor and box is \(\mu_{\mathrm{k}}\) (a) In terms of \(\theta, \mu_{\mathrm{k}},\) and \(w\) calculate \(F .\) (b) For \(w=400 \mathrm{N}\) and \(\mu_{\mathrm{k}}=0.25,\) calculate \(F\) for \(\theta\) ranging from \(0^{\circ}\) to \(90^{\circ}\) in increments of \(10^{\circ} .\) Graph \(F\) versus \(\theta\) .(c) From the general expression in part (a), calculate the value of \(\theta\) for which the value of \(F\) , required to maintain constant speed, is a minimum. (Hint: At a point where a function is minimum, what are the first and second derivatives of the function? Here \(F\) is a function of \(\theta . )\) For the special case of \(w=400 \mathrm{N}\) and \(\mu_{\mathrm{k}}=0.25\) evaluate this optimal \(\theta\) and compare your result to the graph you constructed in part (b).

A man pushes on a piano with mass 180 \(\mathrm{kg}\) so that it slides at constant velocity down a ramp that is inclined at \(11.0^{\circ}\) above the horizontal floor. Neglect any friction acting on the piano. Calculate the magnitude of the force applied by the man if he pushes (a) parallel to the incline and (b) parallel to the floor.
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