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 supplied by the system?

Short answer

Question: Calculate the following values for a package dropped on a horizontal conveyor belt with an initial speed of \(v\): a) The time it takes for the package to stop sliding. b) The package's displacement during this time. c) The energy dissipated by friction. d) The total work supplied by the system. Answer: a) The time it takes for the package to stop sliding is given by: \[t = \frac{v}{\mu_{k} g}\] b) The package's displacement during this time is given by: \[s = \frac{v^{2}}{2\mu_{k} g}\] c) The energy dissipated by friction is given by: \[W = -\frac{1}{2}mv^{2}\] d) The total work supplied by the system is given by: \[W_{total} = -\frac{1}{2}mv^{2}\]

Step-by-step solution

Analyzing the Forces Acting on the Package

As the package is dropped on a horizontal conveyor belt, it will experience gravitational force acting downwards and normal force acting upwards due to contact with the conveyor belt. Since the conveyor belt is horizontal, normal force (N) will be equal to gravitational force (mg) in magnitude. The kinetic frictional force (F_k) between the package and the belt will oppose the motion of the package: \[\begin{aligned} F_{k} = -\mu_{k} N = -\mu_{k} mg \end{aligned}\] This frictional force will cause the package to decelerate and eventually stop sliding.

Calculate the Time for the Package to Stop Sliding (a)

We will use the equations of motion for the package under constant deceleration while sliding: Given that the frictional force is the only force causing deceleration, we can use Newton's second law to find the acceleration (a) of the package: \[\begin{aligned} ma = -\mu_{k} mg \end{aligned}\] Then, divide both sides by m, we find the deceleration: \[\begin{aligned} a = -\mu_{k} g \end{aligned}\] Now, we will use the first equation of motion: \[\begin{aligned} v_f = v_i + at \end{aligned}\] Since the package finally comes to rest, \(v_f = 0\). Thus, the equation becomes: \[\begin{aligned} 0 = v - (-\mu_{k} g)t \end{aligned}\] Solving for t: \[\begin{aligned} t = \frac{v}{\mu_{k} g} \end{aligned}\]

Calculate the Package's Displacement (b)

Now that we have the time (t) it takes for the package to stop sliding, we can use the second equation of motion to find displacement (s) during this time: \[\begin{aligned} s = v_i t + \frac{1}{2}a t^{2} \end{aligned}\] Substitute the values for t and a from Steps 2 and 1: \[\begin{aligned} s = v \left( \frac{v}{\mu_{k} g}\right) + \frac{1}{2}(-\mu_{k} g) \left(\frac{v}{\mu_{k} g} \right)^{2} \end{aligned}\] Simplify this expression, s becomes: \[\begin{aligned} s = \frac{v^{2}}{2\mu_{k} g} \end{aligned}\]

Calculate the Energy Dissipated by Friction (c)

Energy dissipated as heat due to friction is equal to the work done by the frictional force. By definition, work done (W) is given by: \[\begin{aligned} W = F_{k} \cdot s \end{aligned}\] Substitute the values for \(F_k\) and s from Steps 1 and 3: \[\begin{aligned} W = -\mu_{k} mg \left( \frac{v^{2}}{2\mu_{k} g}\right) \end{aligned}\] Simplify, and we have: \[\begin{aligned} W = -\frac{1}{2}mv^{2} \end{aligned}\]

Calculate the Total Work Supplied by the System (d)

Since the package is dropped on the conveyor belt, the gravitational potential energy is converted into kinetic energy, and the work done by the system is equal to the change in kinetic energy of the package. Initially, the package has kinetic energy given by: \[\begin{aligned} KE_{i} = \frac{1}{2}mv_{i}^{2} \end{aligned}\] When the package stops sliding, its kinetic energy is zero, \(KE_f = 0\). The change in kinetic energy (ΔKE) is: \[\begin{aligned} \Delta KE = KE_f - KE_i = - KE_i \end{aligned}\] Thus, the total work supplied by the system is: \[\begin{aligned} W_{total} = \Delta KE = - \frac{1}{2}mv^2 \end{aligned}\] To summarize, a) The time it takes for the package to stop sliding: \[\begin{aligned} t = \frac{v}{\mu_{k} g} \end{aligned}\] b) The package's displacement during this time: \[\begin{aligned} s = \frac{v^{2}}{2\mu_{k} g} \end{aligned}\] c) The energy dissipated by friction: \[\begin{aligned} W = -\frac{1}{2}mv^{2} \end{aligned}\] d) The total work supplied by the system: \[\begin{aligned} W_{total} = -\frac{1}{2}mv^{2} \end{aligned}\]

Key concepts

Conveyor Belt Dynamics

Conveyor belt dynamics involve the study of the forces and movements when an object interacts with a moving belt. When a package is dropped on a moving conveyor belt, several forces come into play. The belt is in constant motion, typically at a known speed, denoted by \( v \). As the package makes contact with the belt, it doesn't immediately move at the belt's speed. Instead, it initially slides across the belt.
The forces acting on the package include:

• Gravitational Force: The downward force due to gravity, \( mg \).
• Normal Force: An upward force exerted by the belt, equal in magnitude to the gravitational force due to the level surface.
• Kinetic Friction: The resistance force between the moving package and the belt, represented as \( F_k = \mu_k mg \).

These forces influence how long the package slides before moving with the conveyor belt—that is, until the friction equals the belt speed.

Newton's Laws of Motion

Newton's laws of motion are fundamental in analyzing the movement of the package on the conveyor belt. Specifically, Newton's second law, \( F = ma \), helps us determine the package's acceleration due to friction. If we consider the frictional force as the only horizontal force acting on the package, the equation becomes:\[a = -\mu_k g\]where \( a \) is the deceleration caused by the belt's kinetic friction. This negative sign indicates that the acceleration is opposite the direction of the belt's movement.
Using this deceleration, you can determine how long it takes for the package to stop sliding:\[t = \frac{v}{\mu_k g}\]This formula reveals that the time to stop sliding depends directly on the initial speed of the belt \( v \) and inversely on both the friction coefficient \( \mu_k \) and gravitational acceleration \( g \). Newton's laws enable us to precisely predict these motions under known forces.

Energy Dissipation

Energy dissipation involves the loss of mechanical energy due to friction, usually transformed into heat. As the package interacts with the conveyor belt, the kinetic friction force causes a reduction in the package's kinetic energy. The work done by the frictional force represents the energy loss:\[W = -\frac{1}{2}mv^2\]This negative sign indicates energy loss from the system. The initial kinetic energy of the package, when first contacting the belt, is \( \frac{1}{2}mv^2 \). As the package comes to rest relative to the belt, this kinetic energy is entirely dissipated as heat.
Thus, energy dissipation is crucial for understanding how systems lose energy over time due to frictional forces.