Newton's Second Law of Motion - Revision history

YorickAndeweg at 02:17, 7 June 2019

2019-06-07T02:17:56Z

← Older revision		Revision as of 22:17, 6 June 2019
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	<!--========================================================================================================================================================================================================================================================================================================================================================================================================================================================================Comment to anyone editing this page: This page may be considered redundant; most of its content is covered by the more complete page titled "Newton's Second Law: the Momentum Principle." Consider deleting this page and moving its practice problems and any other useful information to that page. -Yorick, summer editor of 2019 =================================================================================================================================================================================================================================================================================================================================================================================================================================================================-->		<!--========================================================================================================================================================================================================================================================================================================================================================================================================================================================================Comment to anyone editing this page: This page may be considered redundant; most of its content is covered by the more complete page titled "Newton's Second Law: the Momentum Principle." Consider deleting this page and moving its practice problems and any other useful information to that page. -Yorick, summer editor of 2019 =================================================================================================================================================================================================================================================================================================================================================================================================================================================================-->

	This topic covers Newton's Second Law of Motion.		This topic covers Newton's Second Law of Motion.

YorickAndeweg at 02:17, 7 June 2019

2019-06-07T02:17:45Z

← Older revision		Revision as of 22:17, 6 June 2019
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			<!--========================================================================================================================================================================================================================================================================================================================================================================================================================================================================Comment to anyone editing this page: This page may be considered redundant; most of its content is covered by the more complete page titled "Newton's Second Law: the Momentum Principle." Consider deleting this page and moving its practice problems and any other useful information to that page. -Yorick, summer editor of 2019 =================================================================================================================================================================================================================================================================================================================================================================================================================================================================-->

	This topic covers Newton's Second Law of Motion.		This topic covers Newton's Second Law of Motion.

Inge14: /* Difficult */

2015-11-20T20:02:28Z

Difficult

← Older revision		Revision as of 16:02, 20 November 2015
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	Question:A child on a sled has been given a push so that she is moving up a hill. The hill is covered with icy snow and can be treated as frictionless. The child has a mass of 22.7 kg; the sled has a mass of 3.18 kg. What is the acceleration of the child as she moves up the incline? How far will she go before coming to a stop if her speed is 2 m/s when she starts at the bottom of the hill? She holds on tightly and so does not fall off the sled, and the hill makes an angle of 15.5 0 with the horizontal. Problem found from http://www.uwgb.edu/fenclh/problems/dynamics/2D/3/		Question:A child on a sled has been given a push so that she is moving up a hill. The hill is covered with icy snow and can be treated as frictionless. The child has a mass of 22.7 kg; the sled has a mass of 3.18 kg. What is the acceleration of the child as she moves up the incline? How far will she go before coming to a stop if her speed is 2 m/s when she starts at the bottom of the hill? She holds on tightly and so does not fall off the sled, and the hill makes an angle of 15.5 0 with the horizontal. Problem found from http://www.uwgb.edu/fenclh/problems/dynamics/2D/3/

	Answer:[[File:NSL.JPG]]		Answer:[[File:NSL.JPG\|500px]]

	==Connectedness==		==Connectedness==

Inge14: /* Difficult */

2015-11-20T20:01:06Z

Difficult

← Older revision		Revision as of 16:01, 20 November 2015
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	Question:A child on a sled has been given a push so that she is moving up a hill. The hill is covered with icy snow and can be treated as frictionless. The child has a mass of 22.7 kg; the sled has a mass of 3.18 kg. What is the acceleration of the child as she moves up the incline? How far will she go before coming to a stop if her speed is 2 m/s when she starts at the bottom of the hill? She holds on tightly and so does not fall off the sled, and the hill makes an angle of 15.5 0 with the horizontal. Problem found from http://www.uwgb.edu/fenclh/problems/dynamics/2D/3/		Question:A child on a sled has been given a push so that she is moving up a hill. The hill is covered with icy snow and can be treated as frictionless. The child has a mass of 22.7 kg; the sled has a mass of 3.18 kg. What is the acceleration of the child as she moves up the incline? How far will she go before coming to a stop if her speed is 2 m/s when she starts at the bottom of the hill? She holds on tightly and so does not fall off the sled, and the hill makes an angle of 15.5 0 with the horizontal. Problem found from http://www.uwgb.edu/fenclh/problems/dynamics/2D/3/

	Answer:[[File:NSL.~~jpg~~]]		Answer:[[File:NSL.JPG]]

	==Connectedness==		==Connectedness==

Inge14: /* Difficult */

2015-11-20T20:00:29Z

Difficult

← Older revision		Revision as of 16:00, 20 November 2015
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	Question:A child on a sled has been given a push so that she is moving up a hill. The hill is covered with icy snow and can be treated as frictionless. The child has a mass of 22.7 kg; the sled has a mass of 3.18 kg. What is the acceleration of the child as she moves up the incline? How far will she go before coming to a stop if her speed is 2 m/s when she starts at the bottom of the hill? She holds on tightly and so does not fall off the sled, and the hill makes an angle of 15.5 0 with the horizontal. Problem found from http://www.uwgb.edu/fenclh/problems/dynamics/2D/3/		Question:A child on a sled has been given a push so that she is moving up a hill. The hill is covered with icy snow and can be treated as frictionless. The child has a mass of 22.7 kg; the sled has a mass of 3.18 kg. What is the acceleration of the child as she moves up the incline? How far will she go before coming to a stop if her speed is 2 m/s when she starts at the bottom of the hill? She holds on tightly and so does not fall off the sled, and the hill makes an angle of 15.5 0 with the horizontal. Problem found from http://www.uwgb.edu/fenclh/problems/dynamics/2D/3/

	Answer:		Answer:[[File:NSL.jpg]]

	==Connectedness==		==Connectedness==

Inge14 at 19:58, 20 November 2015

2015-11-20T19:58:17Z

← Older revision		Revision as of 15:58, 20 November 2015
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	Answer: To find out if he is still accelerating the net force on the skydiver needs to be found. After drawing a Free Body Diagram it can be seen that he is experiencing a gravitational force of 833N ((9.8m/s^2)*(85kg)) downwards and an air resistance force of 700N upwards. These are not in balance so he is still accelerating with a net force of 133N downwards. Using F=ma, it can be found that at that exact moment in time the skydiver's acceleration is 1.565m/s^2.		Answer: To find out if he is still accelerating the net force on the skydiver needs to be found. After drawing a Free Body Diagram it can be seen that he is experiencing a gravitational force of 833N ((9.8m/s^2)*(85kg)) downwards and an air resistance force of 700N upwards. These are not in balance so he is still accelerating with a net force of 133N downwards. Using F=ma, it can be found that at that exact moment in time the skydiver's acceleration is 1.565m/s^2.
	===Difficult===		===Difficult===
			Question:A child on a sled has been given a push so that she is moving up a hill. The hill is covered with icy snow and can be treated as frictionless. The child has a mass of 22.7 kg; the sled has a mass of 3.18 kg. What is the acceleration of the child as she moves up the incline? How far will she go before coming to a stop if her speed is 2 m/s when she starts at the bottom of the hill? She holds on tightly and so does not fall off the sled, and the hill makes an angle of 15.5 0 with the horizontal. Problem found from http://www.uwgb.edu/fenclh/problems/dynamics/2D/3/

			Answer:

	==Connectedness==		==Connectedness==

Inge14 at 19:52, 20 November 2015

2015-11-20T19:52:24Z

← Older revision		Revision as of 15:52, 20 November 2015
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	Acceleration form:		Acceleration form:
	<math>{\vec{F}_{net} = m}{\vec{a}}</math> where '''F_net''' is the net force on the system in Newtons, '''m''' is the mass of the system in Kg and '''a''' is the acceleration of the system in m/s^2.		<math>{\vec{F}_{net} = m}{\vec{a}}</math> where '''F_net''' is the net force on the system in Newtons, '''m''' is the mass of the system in Kg and '''a''' is the acceleration of the system in m/s^2.

			Gravitational Extension:
			<math>{\vec{F}_{gravity} = G}{\frac{M_{1}M_{2}}{r^2}}</math>

	===A Computational Model===		===A Computational Model===
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	===Simple===		===Simple===
			Question: A box is pushed on a frictionless surface with a force of 10N it accelerates to the right at 2m/s^2. What is the mass of the box?

			Answer: Using the acceleration form of the second law, F=ma: it can be found that the box has a mass of 5kg.

	===Middling===		===Middling===
			Question: An 85kg skydiver is in free-fall. At some time later he experiences an air resistance force of 700N. Is he still accelerating and if so what is this acceleration?

			Answer: To find out if he is still accelerating the net force on the skydiver needs to be found. After drawing a Free Body Diagram it can be seen that he is experiencing a gravitational force of 833N ((9.8m/s^2)*(85kg)) downwards and an air resistance force of 700N upwards. These are not in balance so he is still accelerating with a net force of 133N downwards. Using F=ma, it can be found that at that exact moment in time the skydiver's acceleration is 1.565m/s^2.
	===Difficult===		===Difficult===

Inge14 at 19:38, 20 November 2015

2015-11-20T19:38:02Z

@@ Line 14: / Line 14: @@
 ===A Computational Model===
-How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]
+Glowscript Binary start program: http://www.glowscript.org/#/user/GlowScriptDemos/folder/Examples/program/BinaryStar-VPython
 ==Examples==

Inge14 at 19:31, 20 November 2015

2015-11-20T19:31:55Z

← Older revision		Revision as of 15:31, 20 November 2015
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	Acceleration form:		Acceleration form:
	<math>{vec{F}_{net} = {m}*{vec{a}}</math>		<math>{\vec{F}_{net} = m}{\vec{a}}</math> where '''F_net''' is the net force on the system in Newtons, '''m''' is the mass of the system in Kg and '''a''' is the acceleration of the system in m/s^2.

	===A Computational Model===		===A Computational Model===

Inge14 at 19:24, 20 November 2015

2015-11-20T19:24:28Z

← Older revision		Revision as of 15:24, 20 November 2015
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	Acceleration form:		Acceleration form:
	<math>{vec{F}_{net} = {m}{vec{a}}</math>		<math>{vec{F}_{net} = {m}*{vec{a}}</math>

	===A Computational Model===		===A Computational Model===