Kinetic Friction - Revision history

YorickAndeweg: /* 3. (Middling) */

2019-07-08T22:39:23Z

3. (Middling)

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 ===3. (Middling)===
-A box of mass <math>m</math> sits on an elevator. The coefficient of kinetic friction between it and the floor of the elevator is <math>\mu_k</math>. The elevator accelerates upwards with an acceleration of <math>a_1</math>. How much force does a worker need to apply to accelerate the box forwards at with an acceleration of <math>a_2</math> (relative to the elevator)?
+A box of mass <math>m</math> sits in an elevator. The coefficient of kinetic friction between it and the floor of the elevator is <math>\mu_k</math>. The elevator accelerates upwards with an acceleration of <math>a_1</math>. How much force does a worker need to apply to accelerate the box forwards at with an acceleration of <math>a_2</math> (relative to the elevator)?
-[[File:kineticfrictionelevtor.png]]
+[[File:kineticfrictionelevator.png]]
 ===4. (Difficult)===

YorickAndeweg: /* 3. (Middling) */

2019-07-08T22:32:47Z

3. (Middling)

← Older revision		Revision as of 18:32, 8 July 2019
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	A box of mass <math>m</math> sits on an elevator. The coefficient of kinetic friction between it and the floor of the elevator is <math>\mu_k</math>. The elevator accelerates upwards with an acceleration of <math>a_1</math>. How much force does a worker need to apply to accelerate the box forwards at with an acceleration of <math>a_2</math> (relative to the elevator)?		A box of mass <math>m</math> sits on an elevator. The coefficient of kinetic friction between it and the floor of the elevator is <math>\mu_k</math>. The elevator accelerates upwards with an acceleration of <math>a_1</math>. How much force does a worker need to apply to accelerate the box forwards at with an acceleration of <math>a_2</math> (relative to the elevator)?

			[[File:kineticfrictionelevtor.png]]

	===4. (Difficult)===		===4. (Difficult)===

YorickAndeweg: /* 3. (Middling) */

2019-07-08T22:26:11Z

3. (Middling)

← Older revision		Revision as of 18:26, 8 July 2019
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	===3. (Middling)===		===3. (Middling)===

	A		A box of mass <math>m</math> sits on an elevator. The coefficient of kinetic friction between it and the floor of the elevator is <math>\mu_k</math>. The elevator accelerates upwards with an acceleration of <math>a_1</math>. How much force does a worker need to apply to accelerate the box forwards at with an acceleration of <math>a_2</math> (relative to the elevator)?

	===4. (Difficult)===		===4. (Difficult)===

YorickAndeweg: /* 3. (Middling) */

2019-07-07T20:17:26Z

3. (Middling)

← Older revision		Revision as of 16:17, 7 July 2019
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	===3. (Middling)===		===3. (Middling)===

			A

	===4. (Difficult)===		===4. (Difficult)===

YorickAndeweg: /* 4. (Difficult) */

2019-07-07T20:14:48Z

4. (Difficult)

← Older revision		Revision as of 16:14, 7 July 2019
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	Using the quadratic equation (<math>t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}</math>),		Using the quadratic equation (<math>t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}</math>),

	<math>t =		<math>t = .708, t = 11.5</math>.

			We are interested in the lowest value for t, which is the first time the puck and the end of the sheet have the same position. At this time, the puck passes the end of the sheet. The second solution represents the time at which the sheet overtakes the puck, which would eventually occur if both objects were to continue accelerating as described by the position equations we used. However, in reality, both objects stop accelerating after the puck leaves the sheet because after they lose contact they no longer exert friction forces on each other.

			<b>The puck reaches the end of the sheet after .708s</b>.

			This answer makes sense because it is slightly longer than it would take for the puck to leave the sheet if there were no friction force; if that were the case, one could simply divide 8m by 12m/s to get .66s.

	==Connectedness==		==Connectedness==

YorickAndeweg: /* 4. (Difficult) */

2019-07-07T20:04:19Z

4. (Difficult)

← Older revision		Revision as of 16:04, 7 July 2019
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	For this problem, it is important to realize that the plastic sheet is not fixed in place; as the kinetic friction force slows down the puck, it also accelerates the sheet. To find out when the puck crosses the end of the sheet, let us find expressions for the position of the puck, <math>x_{puck}</math>, as a function of time, as well as the position of the end of the sheet, <math>x_{end of sheet}</math>, as a function of time. That way, we can set their positions equal to each other and solve for <math>t</math>. Let us define <math>t=0</math> as the time when the puck begins to slide across the sheet and <math>x=0</math> as the position where the puck began (the left end of the sheet as it is depicted in the diagram). Let us define the positive direction as the direction of the puck's travel (towards the right as it is depicted in the diagram). To find the expressions, we will need the accelerations of the puck and of the sheet:		For this problem, it is important to realize that the plastic sheet is not fixed in place; as the kinetic friction force slows down the puck, it also accelerates the sheet. To find out when the puck crosses the end of the sheet, let us find expressions for the position of the puck, <math>x_{puck}</math>, as a function of time, as well as the position of the end of the sheet, <math>x_{end of sheet}</math>, as a function of time. That way, we can set their positions equal to each other and solve for <math>t</math>. Let us define <math>t=0</math> as the time when the puck begins to slide across the sheet and <math>x=0</math> as the position where the puck began (the left end of the sheet as it is depicted in the diagram). Let us define the positive direction as the direction of the puck's travel (towards the right as it is depicted in the diagram). To find the expressions, we will need the accelerations of the puck and of the sheet:

	<math>a_{puck} = \frac{~~f_{puck}~~}{m_{puck}} = \frac{- \mu_k * N}{m_{puck}} = \frac{-.15 * .17 * 9.8}{.17} = -1.47</math>m/s<sup>2</sup>.		<math>a_{puck} = \frac{-f_k}{m_{puck}} = \frac{- \mu_k * N}{m_{puck}} = \frac{-.15 * .17 * 9.8}{.17} = -1.47</math>m/s<sup>2</sup>.

	<math>a_{sheet} = \frac{~~f_{sheet}~~}{m_{sheet}} = \frac{\mu_k * N}{m_{sheet}} = \frac{.15 * .17 * 9.8}{.5} = .50</math>m/s<sup>2</sup>.		<math>a_{sheet} = \frac{f_k}{m_{sheet}} = \frac{\mu_k * N}{m_{sheet}} = \frac{.15 * .17 * 9.8}{.5} = .50</math>m/s<sup>2</sup>.

	To find the expressions for the objects' positions, let us use [[Kinematics]]:		To find the expressions for the objects' positions, let us use [[Kinematics]]:

YorickAndeweg: /* 4. (Difficult) */

2019-07-07T20:03:34Z

4. (Difficult)

← Older revision		Revision as of 16:03, 7 July 2019
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	<math>a_{puck} = \frac{f_{puck}}{m_{puck}} = \frac{- \mu_k * N}{m_{puck}} = \frac{-.15 * .17 * 9.8}{.17} = -1.47</math>m/s<sup>2</sup>.		<math>a_{puck} = \frac{f_{puck}}{m_{puck}} = \frac{- \mu_k * N}{m_{puck}} = \frac{-.15 * .17 * 9.8}{.17} = -1.47</math>m/s<sup>2</sup>.

	<math>x_{~~puck~~}~~(t)~~ = ~~v_0 t +~~ \frac{1}{2}~~at^2 + x_0~~ = ~~asdfasdfasdf~~</math>		<math>a_{sheet} = \frac{f_{sheet}}{m_{sheet}} = \frac{\mu_k * N}{m_{sheet}} = \frac{.15 * .17 * 9.8}{.5} = .50</math>m/s<sup>2</sup>.

	<math>x_{end of sheet}(t) = ~~sadfasdfas~~</math>		To find the expressions for the objects' positions, let us use [[Kinematics]]:

			<math>x_{puck}(t) = x_0 + v_0 t + \frac{1}{2}at^2 = 12t - .735t^2</math>m

			<math>x_{end of sheet}(t) = x_0 + v_0 t + \frac{1}{2}at^2 = 8 + .25t^2</math>m

			Now let us set the expressions for the objects' positions equal to each other to see when the puck crosses the end of the sheet:

			<math>12t - .735t^2 = 8 + .25t^2</math>

			<math>0 = .985t^2 - 12t + 8</math>

			Using the quadratic equation (<math>t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}</math>),

			<math>t =

	==Connectedness==		==Connectedness==

YorickAndeweg: /* 1. (Simple) */

2019-07-07T19:42:37Z

1. (Simple)

← Older revision		Revision as of 15:42, 7 July 2019
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	The magnitude of the acceleration of the box is therefore		The magnitude of the acceleration of the box is therefore

	<math>a = \frac{f}{m} = \frac{17.64}{6} = 2.94</math>m/s^2 by Newton's second law		<math>a = \frac{f}{m} = \frac{17.64}{6} = 2.94</math>m/s<sup>2</sup> by Newton's second law

	We can use this acceleration along with [[Kinematics]] to answer each question. Let us define the positive direction as the direction of the box's velocity.		We can use this acceleration along with [[Kinematics]] to answer each question. Let us define the positive direction as the direction of the box's velocity.

YorickAndeweg: /* 4. (Difficult) */

2019-07-07T19:42:25Z

4. (Difficult)

← Older revision		Revision as of 15:42, 7 July 2019
Line 79:		Line 79:
	For this problem, it is important to realize that the plastic sheet is not fixed in place; as the kinetic friction force slows down the puck, it also accelerates the sheet. To find out when the puck crosses the end of the sheet, let us find expressions for the position of the puck, <math>x_{puck}</math>, as a function of time, as well as the position of the end of the sheet, <math>x_{end of sheet}</math>, as a function of time. That way, we can set their positions equal to each other and solve for <math>t</math>. Let us define <math>t=0</math> as the time when the puck begins to slide across the sheet and <math>x=0</math> as the position where the puck began (the left end of the sheet as it is depicted in the diagram). Let us define the positive direction as the direction of the puck's travel (towards the right as it is depicted in the diagram). To find the expressions, we will need the accelerations of the puck and of the sheet:		For this problem, it is important to realize that the plastic sheet is not fixed in place; as the kinetic friction force slows down the puck, it also accelerates the sheet. To find out when the puck crosses the end of the sheet, let us find expressions for the position of the puck, <math>x_{puck}</math>, as a function of time, as well as the position of the end of the sheet, <math>x_{end of sheet}</math>, as a function of time. That way, we can set their positions equal to each other and solve for <math>t</math>. Let us define <math>t=0</math> as the time when the puck begins to slide across the sheet and <math>x=0</math> as the position where the puck began (the left end of the sheet as it is depicted in the diagram). Let us define the positive direction as the direction of the puck's travel (towards the right as it is depicted in the diagram). To find the expressions, we will need the accelerations of the puck and of the sheet:

	<math>a_{puck} = \frac{f_{puck}}{m_{puck}} = \frac{- \mu_k * N}{m_{puck}} = \frac{-.15 * .17 * 9.8}{.17} = -1.47</math>m/s^2.		<math>a_{puck} = \frac{f_{puck}}{m_{puck}} = \frac{- \mu_k * N}{m_{puck}} = \frac{-.15 * .17 * 9.8}{.17} = -1.47</math>m/s<sup>2</sup>.

	<math>x_{puck}(t) = v_0 t + \frac{1}{2}at^2 + x_0 = asdfasdfasdf</math>		<math>x_{puck}(t) = v_0 t + \frac{1}{2}at^2 + x_0 = asdfasdfasdf</math>

YorickAndeweg: /* 1. (Simple) */

2019-07-07T19:41:59Z

1. (Simple)

← Older revision		Revision as of 15:41, 7 July 2019
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	The magnitude of the acceleration of the box is therefore		The magnitude of the acceleration of the box is therefore

	<math>a = \frac{f}{m} = \frac{17.64}{6} = 2.94</math>m/s by Newton's second law		<math>a = \frac{f}{m} = \frac{17.64}{6} = 2.94</math>m/s^2 by Newton's second law

	We can use this acceleration along with [[Kinematics]] to answer each question. Let us define the positive direction as the direction of the box's velocity.		We can use this acceleration along with [[Kinematics]] to answer each question. Let us define the positive direction as the direction of the box's velocity.