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| <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> |
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| <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.1.2.$ 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. |
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| <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 |
| $$ l = v_{av}t = \frac{v_0 + 0}{2}t = \frac{v_0}{2}t $$ | | $$ l = v_{av}t = \frac{v_0 + 0}{2}t = \frac{v_0}{2}t $$ |
| Acceleration of the body | | Acceleration of the body |
| $$ 0 = v_0 - at\Rightarrow v_0 = at $$ | | $$ 0 = v_0 - at\Rightarrow v_0 = at $$ |
| Then | | Then |
| $$ l = \frac{at}{2}t = \frac{at^2}{2}\Rightarrow a = \frac{2l}{t^2} $$ | | $$ l = \frac{at}{2}t = \frac{at^2}{2}\Rightarrow a = \frac{2l}{t^2} $$ |
| Making a substitution in (1) | | Making a substitution in (1) |
| $$ F_{fr} = \frac{2ml}{t^2} $$ | | $$ F_{fr} = \frac{2ml}{t^2} $$ |
| Calculations | | Calculations |
| $$ F_{fr} = \frac{2 \cdot 0,1 kg \cdot 20 m}{(5 с)^2} = 0.16 \text{ N} $$ | | $$ F_{fr} = \frac{2 \cdot 0,1 kg \cdot 20 m}{(5 с)^2} = 0.16 \text{ N} $$ |
| 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 |
| $$ l = \frac{at^2}{2}\text{, and }a = \frac{2l}{t^2} $$ | | $$ l = \frac{at^2}{2}\text{, and }a = \frac{2l}{t^2} $$ |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$\fbox{$F_{fr} = \frac{2ml}{t^2}$; $F_{fr} = 0.16 \text{ N}$}$$ | | $$\fbox{$F_{fr} = \frac{2ml}{t^2}$; $F_{fr} = 0.16 \text{ N}$}$$ |
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