17. Energins bevarande - Versionshistorik

Tek: flyttade 17. Conservation of energy till 17. Energins bevarande

Tek — Mon, 01 Mar 2010 14:12:17 GMT

flyttade 17. Conservation of energy till 17. Energins bevarande

← Äldre version

Versionen från 1 mars 2010 kl. 14.12

Ian den 18 februari 2010 kl. 18.23

Ian — Thu, 18 Feb 2010 18:23:28 GMT

← Äldre version		Versionen från 18 februari 2010 kl. 18.23
Rad 8:		Rad 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
Rad 23:		Rad 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.	+


Rad 53:		Rad 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}
Rad 71:		Rad 61:
	& =0\textrm{.}07\text{ J}		& =0\textrm{.}07\text{ J}
	\end{align}</math>		\end{align}</math>
-


Rad 78:		Rad 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'''
Rad 90:		Rad 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}} \\
Rad 97:		Rad 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}
Rad 113:		Rad 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}
Rad 122:		Rad 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}
Rad 131:		Rad 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}
Rad 143:		Rad 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}
Rad 156:		Rad 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 den 30 september 2009 kl. 14.19

Ian — Wed, 30 Sep 2009 14:19:45 GMT

← Äldre version		Versionen från 30 september 2009 kl. 14.19
Rad 33:		Rad 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>


Rad 39:		Rad 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>


Rad 55:		Rad 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>


Rad 61:		Rad 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>


Rad 68:		Rad 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]]'''
Rad 91:		Rad 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}
Rad 101:		Rad 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>
Rad 109:		Rad 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>

Rad 118:		Rad 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>

Rad 126:		Rad 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>
Rad 134:		Rad 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>

Rad 150:		Rad 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>
Rad 157:		Rad 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 den 30 september 2009 kl. 14.14

Ian — Wed, 30 Sep 2009 14:14:27 GMT

← Äldre version		Versionen från 30 september 2009 kl. 14.14
Rad 19:		Rad 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>.
Rad 27:		Rad 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>.
Rad 42:		Rad 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:

Rad 67:		Rad 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.
Rad 74:		Rad 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.

Rad 134:		Rad 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 den 30 september 2009 kl. 14.05

Ian — Wed, 30 Sep 2009 14:05:30 GMT

← Äldre version		Versionen från 30 september 2009 kl. 14.05
Rad 21:		Rad 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.
Rad 29:		Rad 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>.


Rad 35:		Rad 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>.


Rad 43:		Rad 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
Rad 70:		Rad 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.

Ian den 30 september 2009 kl. 13.56

Ian — Wed, 30 Sep 2009 13:56:51 GMT

← Äldre version		Versionen från 30 september 2009 kl. 13.56
Rad 6:		Rad 6:
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}
		+
		+
		+	Key Results
		+
		+	Kinetic Energy
		+	<math>\frac{1}{2}m{{v}^{2}}</math>
		+
		+
		+	Potential Energy mgh
		+
		+	Energy lost due to friction = Work Done Against Friction
		+
		+
		+	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) 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?
		+
		+	Solution
		+	(a)
		+	The kinetic energy is at a maximum when the velocity is 16 ms-1.
		+
		+
		+	<math>\text{KE}=\frac{1}{2}\times 0.3\times {{16}^{2}}=38.4\text{ J}</math>
		+
		+
		+	The kinetic energy is at a minimum when the velocity is 6 ms-1.
		+
		+
		+	<math>\text{KE}=\frac{1}{2}\times 0.3\times {{6}^{2}}=5.4\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.
		+	Example 17.2
		+	A ball, of mass 20 grams, hits a wall travelling at 4 ms-1 and rebounds at
		+	3 ms-1. Find the energy lost during the bounce.
		+
		+	Solution
		+	Find the kinetic energy before the bounce:
		+
		+
		+	<math>\text{KE}=\frac{1}{2}\times 0.02\times {{4}^{2}}=0.16\text{ J}</math>
		+
		+
		+	Find the kinetic energy after the bounce:
		+
		+
		+	<math>\text{KE}=\frac{1}{2}\times 0.02\times {{3}^{2}}=0.09\text{ J}</math>
		+
		+
		+	The loss of energy can then be calculated:
		+
		+
		+	<math>\begin{align}
		+	& \text{Energy Lost }=0.16-0.09 \\
		+	& =0.07\text{ J}
		+	\end{align}</math>
		+
		+
		+	Example 17.3
		+	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.
		+
		+
		+
		+
		+
		+
		+
		+
		+
		+
		+	Solution
		+	Find the total energy of the carriage at A.
		+
		+
		+	<math>\begin{align}
		+	& \text{Total Energy}=mgh+\frac{1}{2}m{{v}^{2}} \\
		+	& =500\times 9.8\times 1+\frac{1}{2}\times 500\times {{12}^{2}} \\
		+	& =4900+36000 \\
		+	& =40900\text{ J}
		+	\end{align}</math>
		+
		+
		+	At the point B:
		+
		+
		+	<math>\begin{align}
		+	& 40900=500\times 9.8\times 5+\text{KE} \\
		+	& \text{KE}=\text{4090}0-\text{24500}=16400\text{ J} \\
		+	\end{align}</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& 16400=\frac{1}{2}\times 500{{v}^{2}} \\
		+	& v=\sqrt{\frac{16400}{250}}=8.1\text{ m}{{\text{s}}^{\text{-1}}} \\
		+	\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.
		+
		+
		+	<math>\begin{align}
		+	& 40900=\frac{1}{2}\times 500{{v}^{2}} \\
		+	& v=\sqrt{\frac{40900}{250}}=12.8\text{ m}{{\text{s}}^{\text{-1}}} \\
		+	\end{align}</math>
		+
		+
		+	At the point D:
		+
		+
		+	<math>\begin{align}
		+	& 40900=500\times 9.8\times 3+\text{KE} \\
		+	& \text{KE}=\text{4090}0-\text{14700}=26200\text{ J} \\
		+	\end{align}</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& 26200=\frac{1}{2}\times 500{{v}^{2}} \\
		+	& v=\sqrt{\frac{26200}{250}}=10.2\text{ m}{{\text{s}}^{\text{-1}}} \\
		+	\end{align}</math>
		+
		+
		+	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) 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) 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}
		+	& 60\times 9.8\times 40=\frac{1}{2}\times 60{{v}^{2}} \\
		+	& v=\sqrt{\frac{23520}{30}}=28\text{ m}{{\text{s}}^{\text{-1}}} \\
		+	\end{align}</math>
		+
		+
		+	(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>

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Ian — Wed, 30 Sep 2009 13:51:56 GMT

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