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?

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.

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: \[\begin{aligned} t = \frac{v}{\mu_{k} g} \end{aligned}\]

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: \[\begin{aligned} s = \frac{v^{2}}{2\mu_{k} g} \end{aligned}\]

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: \[\begin{aligned} W = -\frac{1}{2}mv^{2} \end{aligned}\]

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: \[\begin{aligned} W_{total} = -\frac{1}{2}mv^{2} \end{aligned}\]