17. Conservation of energy - Revision history

Ian at 13:11, 27 April 2011

2011-04-27T13:11:32Z

←Older revision		Revision as of 13:11, 27 April 2011
Line 4:		Line 4:
	{{Selected tab\|[[17. Conservation of energy\|Theory]]}}		{{Selected tab\|[[17. Conservation of energy\|Theory]]}}
	{{Not selected tab\|[[17. Exercises\|Exercises]]}}		{{Not selected tab\|[[17. Exercises\|Exercises]]}}
		+	{{Not selected tab\|[[17. Video\|Video]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}
-

	== '''Key Points''' ==		== '''Key Points''' ==

Ian at 11:04, 2 October 2010

2010-10-02T11:04:26Z

←Older revision		Revision as of 11:04, 2 October 2010
Line 26:		Line 26:
	<math>Fd</math>		<math>Fd</math>

-	Thus if there is no friction or friction is negligible the particle does not lose energy~~, in~~ other words its energy is constant (conserved).	+	Thus if there is no friction or friction is negligible the particle does not lose energy. In other words its energy is constant (conserved).

-	In the above any force which resists motion is to be regarded as a frictional force.	+	In the above, any force which resists motion is to be regarded as a frictional force.

Ian at 09:07, 30 September 2010

2010-09-30T09:07:45Z

←Older revision		Revision as of 09:07, 30 September 2010
Line 15:		Line 15:
	<math>\ \frac{1}{2}m{{v}^{2}}</math>		<math>\ \frac{1}{2}m{{v}^{2}}</math>

-	Potential Energy <math>\ mgh</math>	+	Potential Energy: <math>\ mgh</math>

	Total energy is the sum of the potential and kinetic energy of the particle.		Total energy is the sum of the potential and kinetic energy of the particle.

-	Energy lost due to friction = Work ~~Done Against Friction~~	+	Energy lost due to friction = Work done against friction
		+
		+	If <math>F</math> is a constant friction force and <math>d</math> is the distance the particle moves,
		+
		+	Work done against friction:
		+	<math>Fd</math>

	Thus if there is no friction or friction is negligible the particle does not lose energy, in other words its energy is constant (conserved).		Thus if there is no friction or friction is negligible the particle does not lose energy, in other words its energy is constant (conserved).

Ian at 08:56, 30 September 2010

2010-09-30T08:56:51Z

←Older revision		Revision as of 08:56, 30 September 2010
Line 22:		Line 22:

	Thus if there is no friction or friction is negligible the particle does not lose energy, in other words its energy is constant (conserved).		Thus if there is no friction or friction is negligible the particle does not lose energy, in other words its energy is constant (conserved).
		+
		+	In the above any force which resists motion is to be regarded as a frictional force.

Ian at 08:25, 30 September 2010

2010-09-30T08:25:46Z

←Older revision		Revision as of 08:25, 30 September 2010
Line 21:		Line 21:
	Energy lost due to friction = Work Done Against Friction		Energy lost due to friction = Work Done Against Friction

-	Thus if ~~friction does~~ no ~~work~~ or ~~it can be assumed~~ friction is negligible the particle does not lose energy.	+	Thus if there is no friction or friction is negligible the particle does not lose energy, in other words its energy is constant (conserved).

Ian at 16:15, 8 March 2010

2010-03-08T16:15:04Z

←Older revision		Revision as of 16:15, 8 March 2010
Line 9:		Line 9:

	== '''Key Points''' ==		== '''Key Points''' ==
		+
		+	The following theory is for a particle of mass <math>m</math>, having a speed <math>v</math> and at a height <math>h</math> above the ground.

	Kinetic Energy:		Kinetic Energy:
Line 14:		Line 16:

	Potential Energy <math>\ mgh</math>		Potential Energy <math>\ mgh</math>
		+
		+	Total energy is the sum of the potential and kinetic energy of the particle.

	Energy lost due to friction = Work Done Against Friction		Energy lost due to friction = Work Done Against Friction
		+
		+	Thus if friction does no work or it can be assumed friction is negligible the particle does not lose energy.
		+
		+

Ian at 18:23, 18 February 2010

2010-02-18T18:23:28Z

←Older revision		Revision as of 18:23, 18 February 2010
Line 8:		Line 8:


-	Key ~~Results~~	+	== '''Key Points''' ==

-	Kinetic Energy	+	Kinetic Energy:
-	<math>\frac{1}{2}m{{v}^{2}}</math>	+	<math>\ \frac{1}{2}m{{v}^{2}}</math>

-		+	Potential Energy <math>\ mgh</math>
-	Potential Energy mgh	+

	Energy lost due to friction = Work Done Against Friction		Energy lost due to friction = Work Done Against Friction
Line 23:		Line 22:
	A ball, of mass 300 grams, is hit from ground level, so that its initial speed is 16 <math>\text{m}{{\text{s}}^{-1}}</math>. At its maximum height the ball travels at 6 <math>\text{m}{{\text{s}}^{-1}}</math>.		A ball, of mass 300 grams, is hit from ground level, so that its initial speed is 16 <math>\text{m}{{\text{s}}^{-1}}</math>. At its maximum height the ball travels at 6 <math>\text{m}{{\text{s}}^{-1}}</math>.

-	(a) Find the maximum and minimum kinetic energy of the ball.	+	a) Find the maximum and minimum kinetic energy of the ball.

-	(b) If the ball had been projected vertically upwards what would be the minimum kinetic energy of the ball?	+	b) If the ball had been projected vertically upwards what would be the minimum kinetic energy of the ball?

	'''Solution'''		'''Solution'''

-	(a)	+	a)
	The kinetic energy is at a maximum when the velocity is 16 <math>\text{m}{{\text{s}}^{-1}}</math>.		The kinetic energy is at a maximum when the velocity is 16 <math>\text{m}{{\text{s}}^{-1}}</math>.

-
	<math>\text{KE}=\frac{1}{2}\times 0\textrm{.}3\times {{16}^{2}}=38\textrm{.}4\text{ J}</math>		<math>\text{KE}=\frac{1}{2}\times 0\textrm{.}3\times {{16}^{2}}=38\textrm{.}4\text{ J}</math>
-

	The kinetic energy is at a minimum when the velocity is 6 <math>\text{m}{{\text{s}}^{-1}}</math>.		The kinetic energy is at a minimum when the velocity is 6 <math>\text{m}{{\text{s}}^{-1}}</math>.
-

	<math>\text{KE}=\frac{1}{2}\times 0\textrm{.}3\times {{6}^{2}}=5\textrm{.}4\text{ J}</math>		<math>\text{KE}=\frac{1}{2}\times 0\textrm{.}3\times {{6}^{2}}=5\textrm{.}4\text{ J}</math>

-		+	b) If the ball was projected vertically, it would have a velocity of zero at its maximum height and hence the minimum kinetic energy would be zero.
-	(b) If the ball was projected vertically, it would have a velocity of zero at its maximum height and hence the minimum kinetic energy would be zero.	+


Line 53:		Line 48:

	Find the kinetic energy before the bounce:		Find the kinetic energy before the bounce:
-

	<math>\text{KE}=\frac{1}{2}\times 0\textrm{.}02\times {{4}^{2}}=0\textrm{.}16\text{ J}</math>		<math>\text{KE}=\frac{1}{2}\times 0\textrm{.}02\times {{4}^{2}}=0\textrm{.}16\text{ J}</math>
-

	Find the kinetic energy after the bounce:		Find the kinetic energy after the bounce:
-		+
-		+
	<math>\text{KE}=\frac{1}{2}\times 0\textrm{.}02\times {{3}^{2}}=0\textrm{.}09\text{ J}</math>		<math>\text{KE}=\frac{1}{2}\times 0\textrm{.}02\times {{3}^{2}}=0\textrm{.}09\text{ J}</math>
-

	The loss of energy can then be calculated:		The loss of energy can then be calculated:
-

	<math>\begin{align}		<math>\begin{align}
Line 71:		Line 61:
	& =0\textrm{.}07\text{ J}		& =0\textrm{.}07\text{ J}
	\end{align}</math>		\end{align}</math>
-


Line 78:		Line 67:
	A roller coaster has carriages which are modelled as particles of mass		A roller coaster has carriages which are modelled as particles of mass
	500 kg. A carriage starts at A with speed 12 <math>\text{m}{{\text{s}}^{-1}}</math>. It moves, as shown in the diagram, through the points B, C and D. Find the kinetic energy and speed of the carriage at each of these points.		500 kg. A carriage starts at A with speed 12 <math>\text{m}{{\text{s}}^{-1}}</math>. It moves, as shown in the diagram, through the points B, C and D. Find the kinetic energy and speed of the carriage at each of these points.
-
-

	[[Image:E17.3.GIF]]		[[Image:E17.3.GIF]]
-
-
-

	'''Solution'''		'''Solution'''
Line 90:		Line 74:
	Find the total energy of the carriage at A.		Find the total energy of the carriage at A.

-
	<math>\begin{align}		<math>\begin{align}
	& \text{Total Energy}=mgh+\frac{1}{2}m{{v}^{2}} \\		& \text{Total Energy}=mgh+\frac{1}{2}m{{v}^{2}} \\
Line 97:		Line 80:
	& =40900\text{ J}		& =40900\text{ J}
	\end{align}</math>		\end{align}</math>
-

	At the point B:		At the point B:

-
	<math>\begin{align}		<math>\begin{align}
	& 40900=500\times 9\textrm{.}8\times 5+\text{KE} \\		& 40900=500\times 9\textrm{.}8\times 5+\text{KE} \\
	& \text{KE}=\text{4090}0-\text{24500}=16400\text{ J} \\		& \text{KE}=\text{4090}0-\text{24500}=16400\text{ J} \\
	\end{align}</math>		\end{align}</math>
-
-

	<math>\begin{align}		<math>\begin{align}
Line 113:		Line 92:
	& v=\sqrt{\frac{16400}{250}}=8\textrm{.}1\text{ m}{{\text{s}}^{\text{-1}}} \\		& v=\sqrt{\frac{16400}{250}}=8\textrm{.}1\text{ m}{{\text{s}}^{\text{-1}}} \\
	\end{align}</math>		\end{align}</math>
-

	At the point C there is no potential energy, so the kinetic energy is 40900 J. The speed can then be calculated.		At the point C there is no potential energy, so the kinetic energy is 40900 J. The speed can then be calculated.
-

	<math>\begin{align}		<math>\begin{align}
Line 122:		Line 99:
	& v=\sqrt{\frac{40900}{250}}=12\textrm{.}8\text{ m}{{\text{s}}^{\text{-1}}} \\		& v=\sqrt{\frac{40900}{250}}=12\textrm{.}8\text{ m}{{\text{s}}^{\text{-1}}} \\
	\end{align}</math>		\end{align}</math>
-

	At the point D:		At the point D:
-

	<math>\begin{align}		<math>\begin{align}
Line 131:		Line 106:
	& \text{KE}=\text{4090}0-\text{14700}=26200\text{ J} \\		& \text{KE}=\text{4090}0-\text{14700}=26200\text{ J} \\
	\end{align}</math>		\end{align}</math>
-
-

	<math>\begin{align}		<math>\begin{align}
Line 143:		Line 116:

	A bungee jumper has mass 60 kg and uses a rope of length 40m. The jumper comes to rest after falling a distance of 60 m.		A bungee jumper has mass 60 kg and uses a rope of length 40m. The jumper comes to rest after falling a distance of 60 m.
-	(a) Find the speed of the jumper when the rope first becomes taut.
-	(b) Find the elastic potential energy stored in the rope, when it is fully stretched.

-	~~'''Solution'''~~	+	a) Find the speed of the jumper when the rope first becomes taut.

-	(a) ~~When~~ the ~~rope becomes taut the bungee jumper has fallen 40 m. The~~ potential energy ~~lost will be equal to~~ the ~~kinetic energy~~.	+	b) Find the elastic potential energy stored in the rope, when it is fully stretched.

		+	'''Solution'''
		+
		+	a) When the rope becomes taut the bungee jumper has fallen 40 m. The potential energy lost will be equal to the kinetic energy.

	<math>\begin{align}		<math>\begin{align}
Line 156:		Line 130:
	\end{align}</math>		\end{align}</math>

-		+	b) At the lowest point all of the potential energy lost will be equal to the elastic potential energy.
-	(b) At the lowest point all of the potential energy lost will be equal to the elastic potential energy.	+

	<math>\text{EPE }=60\times 9\textrm{.}8\times 60=35280\text{ J}</math>		<math>\text{EPE }=60\times 9\textrm{.}8\times 60=35280\text{ J}</math>

Ian at 14:19, 30 September 2009

2009-09-30T14:19:45Z

←Older revision		Revision as of 14:19, 30 September 2009
Line 33:		Line 33:


-	<math>\text{KE}=\frac{1}{2}\times 0.3\times {{16}^{2}}=38.4\text{ J}</math>	+	<math>\text{KE}=\frac{1}{2}\times 0\textrm{.}3\times {{16}^{2}}=38\textrm{.}4\text{ J}</math>


Line 39:		Line 39:


-	<math>\text{KE}=\frac{1}{2}\times 0.3\times {{6}^{2}}=5.4\text{ J}</math>	+	<math>\text{KE}=\frac{1}{2}\times 0\textrm{.}3\times {{6}^{2}}=5\textrm{.}4\text{ J}</math>


Line 55:		Line 55:


-	<math>\text{KE}=\frac{1}{2}\times 0.02\times {{4}^{2}}=0.16\text{ J}</math>	+	<math>\text{KE}=\frac{1}{2}\times 0\textrm{.}02\times {{4}^{2}}=0\textrm{.}16\text{ J}</math>


Line 61:		Line 61:


-	<math>\text{KE}=\frac{1}{2}\times 0.02\times {{3}^{2}}=0.09\text{ J}</math>	+	<math>\text{KE}=\frac{1}{2}\times 0\textrm{.}02\times {{3}^{2}}=0\textrm{.}09\text{ J}</math>


Line 68:		Line 68:

	<math>\begin{align}		<math>\begin{align}
-	& \text{Energy Lost }=0.16-0.09 \\	+	& \text{Energy Lost }=0\textrm{.}16-0\textrm{.}09 \\
-	& =0.07\text{ J}	+	& =0\textrm{.}07\text{ J}
	\end{align}</math>		\end{align}</math>
		+
		+

	'''[[Example 17.3]]'''		'''[[Example 17.3]]'''
Line 91:		Line 93:
	<math>\begin{align}		<math>\begin{align}
	& \text{Total Energy}=mgh+\frac{1}{2}m{{v}^{2}} \\		& \text{Total Energy}=mgh+\frac{1}{2}m{{v}^{2}} \\
-	& =500\times 9.8\times 1+\frac{1}{2}\times 500\times {{12}^{2}} \\	+	& =500\times 9\textrm{.}8\times 1+\frac{1}{2}\times 500\times {{12}^{2}} \\
	& =4900+36000 \\		& =4900+36000 \\
	& =40900\text{ J}		& =40900\text{ J}
Line 101:		Line 103:

	<math>\begin{align}		<math>\begin{align}
-	& 40900=500\times 9.8\times 5+\text{KE} \\	+	& 40900=500\times 9\textrm{.}8\times 5+\text{KE} \\
	& \text{KE}=\text{4090}0-\text{24500}=16400\text{ J} \\		& \text{KE}=\text{4090}0-\text{24500}=16400\text{ J} \\
	\end{align}</math>		\end{align}</math>
Line 109:		Line 111:
	<math>\begin{align}		<math>\begin{align}
	& 16400=\frac{1}{2}\times 500{{v}^{2}} \\		& 16400=\frac{1}{2}\times 500{{v}^{2}} \\
-	& v=\sqrt{\frac{16400}{250}}=8.1\text{ m}{{\text{s}}^{\text{-1}}} \\	+	& v=\sqrt{\frac{16400}{250}}=8\textrm{.}1\text{ m}{{\text{s}}^{\text{-1}}} \\
	\end{align}</math>		\end{align}</math>

Line 118:		Line 120:
	<math>\begin{align}		<math>\begin{align}
	& 40900=\frac{1}{2}\times 500{{v}^{2}} \\		& 40900=\frac{1}{2}\times 500{{v}^{2}} \\
-	& v=\sqrt{\frac{40900}{250}}=12.8\text{ m}{{\text{s}}^{\text{-1}}} \\	+	& v=\sqrt{\frac{40900}{250}}=12\textrm{.}8\text{ m}{{\text{s}}^{\text{-1}}} \\
	\end{align}</math>		\end{align}</math>

Line 126:		Line 128:

	<math>\begin{align}		<math>\begin{align}
-	& 40900=500\times 9.8\times 3+\text{KE} \\	+	& 40900=500\times 9\textrm{.}8\times 3+\text{KE} \\
	& \text{KE}=\text{4090}0-\text{14700}=26200\text{ J} \\		& \text{KE}=\text{4090}0-\text{14700}=26200\text{ J} \\
	\end{align}</math>		\end{align}</math>
Line 134:		Line 136:
	<math>\begin{align}		<math>\begin{align}
	& 26200=\frac{1}{2}\times 500{{v}^{2}} \\		& 26200=\frac{1}{2}\times 500{{v}^{2}} \\
-	& v=\sqrt{\frac{26200}{250}}=10.2\text{ m}{{\text{s}}^{\text{-1}}} \\	+	& v=\sqrt{\frac{26200}{250}}=10\textrm{.}2\text{ m}{{\text{s}}^{\text{-1}}} \\
	\end{align}</math>		\end{align}</math>

Line 150:		Line 152:

	<math>\begin{align}		<math>\begin{align}
-	& 60\times 9.8\times 40=\frac{1}{2}\times 60{{v}^{2}} \\	+	& 60\times 9\textrm{.}8\times 40=\frac{1}{2}\times 60{{v}^{2}} \\
	& v=\sqrt{\frac{23520}{30}}=28\text{ m}{{\text{s}}^{\text{-1}}} \\		& v=\sqrt{\frac{23520}{30}}=28\text{ m}{{\text{s}}^{\text{-1}}} \\
	\end{align}</math>		\end{align}</math>
Line 157:		Line 159:
	(b) At the lowest point all of the potential energy lost will be equal to the elastic potential energy.		(b) At the lowest point all of the potential energy lost will be equal to the elastic potential energy.

-	<math>\text{EPE }=60\times 9.8\times 60=35280\text{ J}</math>	+	<math>\text{EPE }=60\times 9\textrm{.}8\times 60=35280\text{ J}</math>

Ian at 14:14, 30 September 2009

2009-09-30T14:14:27Z

←Older revision		Revision as of 14:14, 30 September 2009
Line 19:		Line 19:


-	Example 17.1	+	'''[[Example 17.1]]'''

	A ball, of mass 300 grams, is hit from ground level, so that its initial speed is 16 <math>\text{m}{{\text{s}}^{-1}}</math>. At its maximum height the ball travels at 6 <math>\text{m}{{\text{s}}^{-1}}</math>.		A ball, of mass 300 grams, is hit from ground level, so that its initial speed is 16 <math>\text{m}{{\text{s}}^{-1}}</math>. At its maximum height the ball travels at 6 <math>\text{m}{{\text{s}}^{-1}}</math>.
Line 27:		Line 27:
	(b) If the ball had been projected vertically upwards what would be the minimum kinetic energy of the ball?		(b) If the ball had been projected vertically upwards what would be the minimum kinetic energy of the ball?

-	Solution	+	'''Solution'''
		+
	(a)		(a)
	The kinetic energy is at a maximum when the velocity is 16 <math>\text{m}{{\text{s}}^{-1}}</math>.		The kinetic energy is at a maximum when the velocity is 16 <math>\text{m}{{\text{s}}^{-1}}</math>.
Line 42:		Line 43:

	(b) If the ball was projected vertically, it would have a velocity of zero at its maximum height and hence the minimum kinetic energy would be zero.		(b) If the ball was projected vertically, it would have a velocity of zero at its maximum height and hence the minimum kinetic energy would be zero.
-	Example 17.2	+
		+
		+	'''[[Example 17.2]]'''
		+
	A ball, of mass 20 grams, hits a wall travelling at 4 <math>\text{m}{{\text{s}}^{-1}}</math> and rebounds at		A ball, of mass 20 grams, hits a wall travelling at 4 <math>\text{m}{{\text{s}}^{-1}}</math> and rebounds at
	3 <math>\text{m}{{\text{s}}^{-1}}</math>. Find the energy lost during the bounce.		3 <math>\text{m}{{\text{s}}^{-1}}</math>. Find the energy lost during the bounce.

-	Solution	+	'''Solution'''
		+
	Find the kinetic energy before the bounce:		Find the kinetic energy before the bounce:

Line 67:		Line 72:
	\end{align}</math>		\end{align}</math>

		+	'''[[Example 17.3]]'''

-	Example 17.3
	A roller coaster has carriages which are modelled as particles of mass		A roller coaster has carriages which are modelled as particles of mass
	500 kg. A carriage starts at A with speed 12 <math>\text{m}{{\text{s}}^{-1}}</math>. It moves, as shown in the diagram, through the points B, C and D. Find the kinetic energy and speed of the carriage at each of these points.		500 kg. A carriage starts at A with speed 12 <math>\text{m}{{\text{s}}^{-1}}</math>. It moves, as shown in the diagram, through the points B, C and D. Find the kinetic energy and speed of the carriage at each of these points.
Line 74:		Line 79:


		+	[[Image:E17.3.GIF]]




		+	'''Solution'''

-
-
-	Solution
	Find the total energy of the carriage at A.		Find the total energy of the carriage at A.

Line 134:		Line 138:


-	Example 17.4	+	'''[[Example 17.4]]'''
		+
	A bungee jumper has mass 60 kg and uses a rope of length 40m. The jumper comes to rest after falling a distance of 60 m.		A bungee jumper has mass 60 kg and uses a rope of length 40m. The jumper comes to rest after falling a distance of 60 m.
	(a) Find the speed of the jumper when the rope first becomes taut.		(a) Find the speed of the jumper when the rope first becomes taut.
	(b) Find the elastic potential energy stored in the rope, when it is fully stretched.		(b) Find the elastic potential energy stored in the rope, when it is fully stretched.

-	Solution	+	'''Solution'''
		+
	(a) When the rope becomes taut the bungee jumper has fallen 40 m. The potential energy lost will be equal to the kinetic energy.		(a) When the rope becomes taut the bungee jumper has fallen 40 m. The potential energy lost will be equal to the kinetic energy.

Ian at 14:05, 30 September 2009

2009-09-30T14:05:30Z

←Older revision		Revision as of 14:05, 30 September 2009
Line 21:		Line 21:
	Example 17.1		Example 17.1

-	A ball, of mass 300 grams, is hit from ground level, so that its initial speed is 16 ms-1. At its maximum height the ball travels at 6 ms-1.	+	A ball, of mass 300 grams, is hit from ground level, so that its initial speed is 16 <math>\text{m}{{\text{s}}^{-1}}</math>. At its maximum height the ball travels at 6 <math>\text{m}{{\text{s}}^{-1}}</math>.

	(a) Find the maximum and minimum kinetic energy of the ball.		(a) Find the maximum and minimum kinetic energy of the ball.
Line 29:		Line 29:
	Solution		Solution
	(a)		(a)
-	The kinetic energy is at a maximum when the velocity is 16 ms-1.	+	The kinetic energy is at a maximum when the velocity is 16 <math>\text{m}{{\text{s}}^{-1}}</math>.


Line 35:		Line 35:


-	The kinetic energy is at a minimum when the velocity is 6 ms-1.	+	The kinetic energy is at a minimum when the velocity is 6 <math>\text{m}{{\text{s}}^{-1}}</math>.


Line 43:		Line 43:
	(b) If the ball was projected vertically, it would have a velocity of zero at its maximum height and hence the minimum kinetic energy would be zero.		(b) If the ball was projected vertically, it would have a velocity of zero at its maximum height and hence the minimum kinetic energy would be zero.
	Example 17.2		Example 17.2
-	A ball, of mass 20 grams, hits a wall travelling at 4 ms-1 and rebounds at	+	A ball, of mass 20 grams, hits a wall travelling at 4 <math>\text{m}{{\text{s}}^{-1}}</math> and rebounds at
-	3 ms-1. Find the energy lost during the bounce.	+	3 <math>\text{m}{{\text{s}}^{-1}}</math>. Find the energy lost during the bounce.

	Solution		Solution
Line 70:		Line 70:
	Example 17.3		Example 17.3
	A roller coaster has carriages which are modelled as particles of mass		A roller coaster has carriages which are modelled as particles of mass
-	500 kg. A carriage starts at A with speed 12 ms-1. It moves, as shown in the diagram, through the points B, C and D. Find the kinetic energy and speed of the carriage at each of these points.	+	500 kg. A carriage starts at A with speed 12 <math>\text{m}{{\text{s}}^{-1}}</math>. It moves, as shown in the diagram, through the points B, C and D. Find the kinetic energy and speed of the carriage at each of these points.