Question

A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of v. The coefficient of kinetic friction between sled and ice is ${\mathit{\mu }}_{\mathbf{k}}$. Use energy considerations to find the distance the sled moves before it stops.

Short answer

The distance of the sled moves before it stops is

$d=\frac{v^2}{2{\mu }_{k}g}$

Step-by-step solution

Definition of the non-isolated system and expression for the energy

A non-isolated system is one for which energy crosses the boundary of the system. An isolated system is one for which no energy crosses the boundary of the system.

If a friction force of magnitude$f_k$ acts over a distance d within a system, the change in internal energy of the system is

$\Delta E_{int}=f_{k}d..........\left(8.14\right)$

Non isolated System (Energy): The most general statement describing the behavior of a non isolated system is the conservation of energy equation

$\Delta E_{system}=\sum T..........\left(8.1\right)$

Including the types of energy storage and energy transfer that we have discussed gives

$\Delta K+\Delta U+\Delta E_{int}=W+Q+T_{MW}+T_{ET}+T_{ER}$

For a specific problem, this equation is generally reduced to a smaller number of terms by eliminating the terms that are not appropriate to the situation.

Finding the expression for the distance

We use the non-isolated system energy model, here written as $-f_{k}d=k_f-k_i$, where the kinetic energy change of the sled after the kick results only from the friction between the sled and ice. From using the equation (8.1) and (8.14)

$$\begin{gathered} \Delta K+\Delta U=-f_{k}d \\ 0-\frac{1}{2}m{v}^2=-f_{k}d \\ \frac{1}{2}m{v}^2={\mu }_{k}mgd..........\left(\because f_k={\mu }_{k}mg\right) \end{gathered}$$

This gives

$d=\frac{v^2}{2{\mu }_{k}g}$

The distance of the sled moves before it stops is

$d=\frac{v^2}{2{\mu }_{k}g}$