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A warehouse worker exerts a constant horizontal force of magnitude \(85 \mathrm{~N}\) on a \(40 \mathrm{~kg}\) box that is initially at rest on the horizontal floor of the warehouse. When the box has moved a distance of \(1.4 \mathrm{~m}\), its speed is \(1.0 \mathrm{~m} / \mathrm{s}\). What is the coefficient of kinetic friction between the box and the floor?

Short Answer

Expert verified
The coefficient of kinetic friction is approximately 0.18.

Step by step solution

01

Identify the variables given

First, list the given variables:- Force exerted by the worker, \( F = 85 \text{ N} \).- Mass of the box, \( m = 40 \text{ kg} \).- Distance moved, \( d = 1.4 \text{ m} \).- Final speed of the box, \( v = 1.0 \text{ m/s} \).- Initial speed of the box, \( u = 0 \text{ m/s} \) (since it starts from rest). We need to find the coefficient of kinetic friction \( \mu_k \).
02

Calculate the kinetic energy change

The initial kinetic energy \( KE_i = \frac{1}{2} m u^2 = 0 \).The final kinetic energy \( KE_f = \frac{1}{2} m v^2 = \frac{1}{2} \times 40 \times (1)^2 = 20 \text{ J} \).The change in kinetic energy \( \Delta KE = KE_f - KE_i = 20 - 0 = 20 \text{ J} \).
03

Determine the work done by the net force

The work-energy principle states that the work done by the net force is equal to the change in kinetic energy.Therefore, the work done \( W = \Delta KE = 20 \text{ J} \).
04

Express net work done as a sum of forces

The net work done is also the result of the applied force and friction force together.The work done by the applied force is \( W_F = F \cdot d = 85 \cdot 1.4 = 119 \text{ J} \).The work done by friction \( W_{friction} = W_F - W_{net} = 119 - 20 = 99 \text{ J} \).
05

Relate friction force to kinetic friction coefficient

The friction force can also be expressed as \( f_k = \mu_k N \), where \( N = m \cdot g \) is the normal force.\( f_k = \mu_k \cdot 40 \cdot 9.8 \).The work done by friction is \( W_{friction} = f_k \cdot d = \mu_k \cdot 40 \cdot 9.8 \cdot 1.4 \).Using the found work done by friction: \( \mu_k \cdot 40 \cdot 9.8 \cdot 1.4 = 99 \).
06

Solve for the coefficient of kinetic friction

Now, solve the equation \( \mu_k \cdot 40 \cdot 9.8 \cdot 1.4 = 99 \).\( \mu_k = \frac{99}{40 \times 9.8 \times 1.4} \).Therefore, \( \mu_k \approx 0.18 \).

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Key Concepts

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

Kinetic Friction
Kinetic friction is the force that opposes the motion when one object slides over another. This type of frictional force acts in the opposite direction to the motion of the object. It depends on two main factors:
  • The nature of the surfaces in contact
  • The normal force pressing the two surfaces together
In the context of the problem, kinetic friction acts to oppose the motion of the box as it is pushed across the warehouse floor. The formula for calculating the force of kinetic friction is given by:
\[ f_k = \mu_k \cdot N \]where:
  • \( f_k \) is the kinetic friction force
  • \( \mu_k \) is the coefficient of kinetic friction
  • \( N \) is the normal force, which is typically the weight of the object if on a horizontal surface \( N = m \cdot g \)
The coefficient of kinetic friction is a dimensionless value that describes how easily one object will slide over another. It varies between different material pairs and needs experimental methods for determination, as calculated in this exercise.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics that relates the net work done on an object to its change in kinetic energy. This principle can be expressed as follows:
\[ W_{net} = \Delta KE \]where:
  • \( W_{net} \) is the net work done on the object
  • \( \Delta KE \) is the change in the object’s kinetic energy
When the worker exerts a force on the box, he does work that changes the box's kinetic energy from zero (since it starts from a rest position) to its final kinetic energy determined by its velocity and mass.
In this exercise, it has been determined that the work completed by the worker (minus the work lost to friction) corresponds precisely to the change in kinetic energy of the box, showing how effectively energy has been transferred from the worker's effort into the motion of the box.
Kinetic Energy Change
Kinetic energy is the energy of an object due to its motion. The formula to calculate kinetic energy is given by:
\[ KE = \frac{1}{2} m v^2 \]where:
  • \( KE \) is the kinetic energy
  • \( m \) is the mass of the object
  • \( v \) is the velocity of the object
The change in kinetic energy \( \Delta KE \) is calculated as the difference between the final kinetic energy \( KE_f \) and the initial kinetic energy \( KE_i \). In this exercise, the initial kinetic energy was zero because the box started from rest. As the box moved under the influence of the worker's force, its kinetic energy increased by 20 J (joules) because of the additional speed it acquired.
Coefficient of Friction
The coefficient of friction is a measure of how much force is required to move one object over another. It is represented by \( \mu \), and can be separated into static (\( \mu_s \)) and kinetic (\( \mu_k \)) coefficients. The kinetic coefficient of friction applies to surfaces in relative motion, like the box sliding on the floor, which is relevant in this exercise.
The formula relating the coefficient of kinetic friction to the frictional force is:\[ \mu_k = \frac{f_k}{N} \]where:
  • \( \mu_k \) is the coefficient of kinetic friction
  • \( f_k \) is the frictional force
  • \( N \) is the normal force
In the scenario given, the coefficient of kinetic friction \( \mu_k \) was calculated to be approximately 0.18, indicating a moderate level of friction between the box and the warehouse floor. This value is critical for understanding how much force is necessary to keep the box moving at a given speed.

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Most popular questions from this chapter

\(68 \mathrm{~kg}\) crate is dragged across a floor by pulling on a rope attached to the crate and inclined \(15^{\circ}\) above the horizontal. (a) If the coefficient of static friction is \(0.50\), what minimum force magnitude is required from the rope to start the crate moving? (b) If \(\mu_{k}=0.35\), what is the magnitude of the initial acceleration of the crate?

A circular curve of highway is designed for traffic moving at \(60 \mathrm{~km} / \mathrm{h}\). Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is \(150 \mathrm{~m}\), what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at \(60 \mathrm{~km} / \mathrm{h} ?\)

A person pushes horizontally with a force of \(220 \mathrm{~N}\) on a \(55 \mathrm{~kg}\) crate to move it across a level floor. The coefficient of kinetic friction between the crate and the floor is \(0.35 .\) What is the magnitude of (a) the frictional force and (b) the acceleration of the crate?

A \(1.5 \mathrm{~kg}\) box is initially at rest on a horizontal surface when at \(t=0\) a horizontal force \(\vec{F}=(1.8 t) \hat{\mathrm{i}} \mathrm{N}\) (with \(t\) in seconds) is applied to the box. The acceleration of the box as a function of time \(t\) is given by \(\vec{a}=0\) for \(0 \leq t \leq 2.8 \mathrm{~s}\) and \(\vec{a}=(1.2 t-2.4) \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}^{2}\) for \(t>\) \(2.8 \mathrm{~s}\). (a) What is the coefficient of static friction between the box and the surface? (b) What is the coefficient of kinetic friction between the box and the surface?

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of \(0.25\) with the floor. If the train is initially moving at a speed of \(48 \mathrm{~km} / \mathrm{h}\), in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

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