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<meta name="description" content="The puck, which was sliding on the ice, stopped after a time t = 5 s after hitting the stick at a distance l = 20 m from the place of impact. The mass of the washer is m = 100 g. Determine the friction force acting on the washer.">
<meta property="og:title" content="The puck, which was sliding on the ice, stopped after a time t = 5 s after hitting the stick at a distance l = 20 m from the place of impact. The mass of the washer is m = 100 g. Determine the friction force acting on the washer.">
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<meta property="og:description" content="The puck, which was sliding on the ice, stopped after a time t = 5 s after hitting the stick at a distance l = 20 m from the place of impact. The mass of the washer is m = 100 g. Determine the friction force acting on the washer.">
<title>The puck, which was sliding on the ice, stopped after a time t = 5 s after hitting the stick at a distance l = 20 m from the place of impact. The mass of the washer is m = 100 g. Determine the friction force acting on the washer.</title>
$2.1.2.$ The puck, which was sliding on the ice, stopped after a time $t = 5$ s after hitting the stick at a distance $l = 20$ m from the place of impact. The mass of the washer is $m = 100$ g. Determine the friction force acting on the washer.
</p>
<h3>Solution</h3>
<p>
<p>According to Newton's second law
$$ m\vec{a} = m\vec{g} + \vec{N} + \vec{F_{fr}}$$
In projection onto the direction of movement
$$ ma = F_{fr}\;(1) $$
To find the friction force, we need to determine the acceleration of the body. We will suggest one of the methods. The distance traveled by the body before stopping is equal to
It is interesting that acceleration can be obtained even more sifrly by using the “method by contradiction” in solving the problem, if from the stopping point, accelerate the puck back to the stick, then
$$ l = \frac{at^2}{2}\text{, and }a = \frac{2l}{t^2}$$
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="description" content="The puck, which was sliding on the ice, stopped after a time t = 5 s after hitting the stick at a distance l = 20 m from the place of impact. The mass of the washer is m = 100 g. Determine the friction force acting on the washer.">
<meta name="description" content="The puck, which was sliding on the ice, stopped after a time t = 5 s after hitting the stick at a distance l = 20 m from the place of impact. The mass of the washer is m = 100 g. Determine the friction force acting on the washer.">
<meta property="og:title" content="The puck, which was sliding on the ice, stopped after a time t = 5 s after hitting the stick at a distance l = 20 m from the place of impact. The mass of the washer is m = 100 g. Determine the friction force acting on the washer.">
<meta property="og:title" content="The puck, which was sliding on the ice, stopped after a time t = 5 s after hitting the stick at a distance l = 20 m from the place of impact. The mass of the washer is m = 100 g. Determine the friction force acting on the washer.">
<meta property="og:image" content="img/logo.png">
<meta property="og:image" content="img/logo.png">
@@ -14,9 +14,9 @@
<meta property="og:description" content="The puck, which was sliding on the ice, stopped after a time t = 5 s after hitting the stick at a distance l = 20 m from the place of impact. The mass of the washer is m = 100 g. Determine the friction force acting on the washer.">
<meta property="og:description" content="The puck, which was sliding on the ice, stopped after a time t = 5 s after hitting the stick at a distance l = 20 m from the place of impact. The mass of the washer is m = 100 g. Determine the friction force acting on the washer.">
<title>The puck, which was sliding on the ice, stopped after a time t = 5 s after hitting the stick at a distance l = 20 m from the place of impact. The mass of the washer is m = 100 g. Determine the friction force acting on the washer.</title>
<title>The puck, which was sliding on the ice, stopped after a time t = 5 s after hitting the stick at a distance l = 20 m from the place of impact. The mass of the washer is m = 100 g. Determine the friction force acting on the washer.</title>
$2.1.2.$ The puck, which was sliding on the ice, stopped after a time $t = 5$ s after hitting the stick at a distance $l = 20$ m from the place of impact. The mass of the washer is $m = 100$ g. Determine the friction force acting on the washer.
$2.1.2.$ The puck, which was sliding on the ice, stopped after a time $t = 5$ s after hitting the stick at a distance $l = 20$ m from the place of impact. The mass of the washer is $m = 100$ g. Determine the friction force acting on the washer.
</p>
</p>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
<p>According to Newton's second law
<p>According to Newton's second law
$$ m\vec{a} = m\vec{g} + \vec{N} + \vec{F_{fr}}$$
$$ m\vec{a} = m\vec{g} + \vec{N} + \vec{F_{fr}}$$
In projection onto the direction of movement
In projection onto the direction of movement
$$ ma = F_{fr}\;(1) $$
$$ ma = F_{fr}\;(1) $$
To find the friction force, we need to determine the acceleration of the body. We will suggest one of the methods. The distance traveled by the body before stopping is equal to
To find the friction force, we need to determine the acceleration of the body. We will suggest one of the methods. The distance traveled by the body before stopping is equal to
It is interesting that acceleration can be obtained even more sifrly by using the “method by contradiction” in solving the problem, if from the stopping point, accelerate the puck back to the stick, then
It is interesting that acceleration can be obtained even more sifrly by using the “method by contradiction” in solving the problem, if from the stopping point, accelerate the puck back to the stick, then
$$ l = \frac{at^2}{2}\text{, and }a = \frac{2l}{t^2}$$
$$ l = \frac{at^2}{2}\text{, and }a = \frac{2l}{t^2}$$
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
