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Expert-verified Found in: Page 237 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of ${\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbit{m}}{\mathbf{/}}{\mathbit{s}}$. The coefficient of kinetic friction between sled and ice is ${\mathbf{0}}{\mathbf{.}}{\mathbf{100}}$. Use energy considerations to find the distance the sled moves before it stops.

The distance the sled moves before it stops is $d=2.04m$

See the step by step solution

## Step 1: Definition of the non-isolated system

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

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

## Step 2: Calculating the distance

We could solve this problem using Newton’s second law, but we will use the non isolated system energy model, from using the equation (8.1) and (8.14) 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.

The weight and normal force both act at to the motion, and therefore do no work on the sled. The friction force is

role="math" localid="1663680345416" ${f}_{k}={\mu }_{k}n={\mu }_{k}mgd$

Since the final kinetic energy is zero, we have

$-{f}_{k}d={k}_{i}\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}_{i}^{2}={\mu }_{k}mg\phantom{\rule{0ex}{0ex}}$

Solving, we get

$d=\frac{m{v}_{i}^{2}}{2{f}_{k}}\phantom{\rule{0ex}{0ex}}=\frac{m{v}_{i}^{2}}{2{\mu }_{k}mg}\phantom{\rule{0ex}{0ex}}=\frac{{v}_{i}^{2}}{2{\mu }_{k}g}\phantom{\rule{0ex}{0ex}}$

Substituting the values, we get

$d=\frac{{\left(2m/s\right)}^{2}}{2\left(0.100\right)\left(9.8m/{s}^{2}\right)}\phantom{\rule{0ex}{0ex}}=2.04m$

Thus, the distance is 2.04 m ### Want to see more solutions like these? 