Question

A package is dropped on a horizontal conveyor belt. The mass of the package is \(m\), the speed of the conveyor belt is \(v\), and the coefficient of kinetic friction between the package and the belt is \(\mu_{\mathrm{k}}\) a) How long does it take for the package to stop sliding on the belt? b) What is the package's displacement during this time? c) What is the energy dissipated by friction? d) What is the total work done by the conveyor belt?

Short answer

Based on the step-by-step solution provided: a) The stopping time of the package is given by \(t = \frac{-v}{\mu_{k} \times g}\). b) The package's displacement while coming to a stop is given by: \(x = v \times \frac{-v}{\mu_{k} \times g} + \frac{1}{2} \times \mu_{k} \times g \times \left(\frac{-v}{\mu_{k} \times g}\right)^2\). c) The energy dissipated by friction is given by: \(E_{dissipated} = \frac{1}{2} \times m \times v^2\). d) The total work done by the conveyor belt is given by: \(W_{total} = \frac{1}{2} \times m \times v^2\).

Step-by-step solution

Determine the stopping time of the package

The package slides due to the frictional force caused by the conveyor belt. According to Newton's second law, the frictional force (\(F_{fr}\)) acting on the package is equal to the mass of the package times its acceleration (\(a\)). \(F_{fr} = m \times a\) The frictional force is also given by: \(F_{fr} = \mu_{k} \times N\) Where \(N = m \times g\) (normal force) represents the downward force acting on the package due to gravity, and \(g\) is the acceleration due to gravity. Thus, we can rewrite the Newton's second law equation as: \(m \times a = \mu_{k} \times m \times g\) Divide both sides by \(m\) and rearrange for acceleration: \(a = \mu_{k} \times g\) As the package is moving at an initial speed \(v\) and comes to rest, the stopping time can be found using the equation: \(t = \frac{v_f - v}{a}\) Where \(v_f = 0\) (final velocity). Now, substitute the equation for acceleration into the stopping time equation: \(t = \frac{-v}{\mu_{k} \times g}\) #a# The stopping time of the package is given by \(t = \frac{-v}{\mu_{k} \times g}\).

Find the package's displacement

To find the package's displacement while it is coming to a stop, we will use the kinematic equation: \(x = v \times t + \frac{1}{2} \times a \times t^2\) We already have the equation for acceleration and stopping time: \(a = \mu_{k} \times g\) \(t = \frac{-v}{\mu_{k} \times g}\) Substitute these equations into the kinematic equation: \(x = v \times \frac{-v}{\mu_{k} \times g} + \frac{1}{2} \times \mu_{k} \times g \times \left(\frac{-v}{\mu_{k} \times g}\right)^2\) #b# The package's displacement while coming to a stop is given by: \(x = v \times \frac{-v}{\mu_{k} \times g} + \frac{1}{2} \times \mu_{k} \times g \times \left(\frac{-v}{\mu_{k} \times g}\right)^2\).

Calculate the energy dissipated by friction

The energy dissipated by friction can be calculated as the difference in kinetic energy. Since the package comes to rest, its final kinetic energy is zero. Initially, it has kinetic energy given by: \(KE_{initial} = \frac{1}{2} \times m \times v^2\) The energy dissipated by friction is equal to the initial kinetic energy: \(E_{dissipated} = KE_{initial} = \frac{1}{2} \times m \times v^2\) #c# The energy dissipated by friction is given by: \(E_{dissipated} = \frac{1}{2} \times m \times v^2\).

Calculate the total work done by the conveyor belt

The total work done by the conveyor belt on the package corresponds to the change in kinetic energy as well as the energy dissipated by friction. Since the initial kinetic energy is equal to the energy dissipated by friction, the total work done is given by: \(W_{total} = E_{dissipated} = \frac{1}{2} \times m \times v^2\) #d# The total work done by the conveyor belt is given by: \(W_{total} = \frac{1}{2} \times m \times v^2\).

Key concepts

Newton's second law

Newton's second law provides the fundamental framework for understanding how objects behave when there is a net force acting on them. It states that the acceleration (\(a\text{ in } \text{m/s}^2\text{)}\) of an object is directly proportional to the net force (\(F\text{ in Newtons, or N}\text{)}\) applied to it and inversely proportional to its mass (\(m\text{ in kilograms, or kg}\text{)}\). Mathematically, we express this law as \(F = m \times a\text{)}.

In the context of the exercise, the net force acting on the package is the frictional force due to its interaction with the conveyor belt. This force causes a deceleration that brings the package to a stop. By calculating the acceleration due to friction, we could then determine how long it takes for the package to stop, which is essential for solving parts a and b of the exercise. It's crucial for students to understand that the same principle applies whether an object is speeding up or slowing down - Newton's second law governs its motion.

Kinematic equations

Kinematic equations are the tools that allow us to describe the motion of objects in terms of displacement (\(x\text{ in meters, or m}\text{)}\), velocity (\(v\text{ in } \text{m/s}\text{)}\), acceleration (\(a\text{ in } \text{m/s}^2\text{)}\), and time (\(t\text{ in seconds, or s}\text{)}\). One of the basic kinematic formulas used in this exercise is \(x = v \times t + \frac{1}{2} \times a \times t^2\text{)}, which relates these variables when acceleration is constant.

For the given problem, the equation helps determine the package's displacement as it decelerates and stops due to kinetic friction. Students should practice using this equation to become comfortable with solving for any of the variables, depending on what information is provided or sought.

Energy dissipation

Energy dissipation refers to the process by which mechanical energy is transformed into other forms of energy, typically heat, due to forces like friction or air resistance. In the case of kinetic friction, the work done by this force on a sliding object is converted into thermal energy, which is effectively 'lost' to the surroundings.

This concept is exemplified in part c of the exercise, where the energy dissipated by friction is equal to the initial kinetic energy of the package. It's important for students to recognize that energy is conserved but can change forms; the mechanical energy of the moving package is not lost but is simply transferred to the environment as heat.

Work-energy principle

The work-energy principle is a fundamental concept in physics that states that the work done on an object results in a change in its kinetic energy. Work (\(W\text{ in joules, or J}\text{)}\) is defined as the force applied to an object times the distance over which the force is applied, and its mathematical expression is \(W = F \times d\text{)}, where \(d\text{)}\ represents distance.

In the context of the exercise, both kinetic energy and work are related to the stopping process of the package on the conveyor belt. Calculating the total work done by the conveyor belt, as shown in the solution for part d, relies directly on the energy dissipated by the frictional force. Students should appreciate that this relationship gives us powerful ways to analyze physical situations using energy considerations, not just forces and acceleration.