Forest UCM Energy KEnWork - Revision history

Foretony: /* Derivation */

2014-09-22T18:48:40Z

Derivation

2014-09-22T15:20:03Z

Problem 4.2

2014-09-22T15:18:58Z

Problem 4.2

2014-09-22T15:17:54Z

Problem 4.2

2014-09-22T15:16:52Z

Problem 4.2

2014-09-22T15:15:38Z

Problem 4.2

2014-09-22T15:14:25Z

Problem 4.2

2014-09-22T03:06:28Z

Line Integral

2014-09-22T03:04:28Z

Line Integral

2014-09-22T02:50:21Z

Line Integral

@@ Line 19: / Line 19: @@
 or
-<math>dT = \vec F_{\mbox{net}} \cdot \vec v dt =\vec F_{\mbox{net}} \cdot d\vec r \equiv d W</math>
+<math>dT = \vec F_{\mbox{net}} \cdot \vec v dt =\vec F_{\mbox{net}} \cdot d\vec r</math>
 or
-:<math>\Delta T = \Delta W</math>
+:<math>\Delta T =\int  \vec F_{\mbox{net}} \cdot d\vec r \equiv W</math>
 ;: The change in a particle's kinetic energy is equivalent to the work done by the net Force used to move the particle

@@ Line 70: / Line 70: @@
 a.)
-<math>W = \left ( \int_0^Q \vec F \cdot d \vec r + \int_Q^P \vec F \cdot d \vec r\right )</math>
+<math>W = \left ( \int_0^P \vec F \cdot dx \hat i + \int_0^P  \vec F \cdot dy \hat j\right )</math>
 :<math>= \left ( \int_0^1 \vec F \cdot dx \hat i + \int_0^1  \vec F \cdot dy \hat j\right )</math>
 :<math>= \left ( \int_0^1 x^2 dx + \int_0^1  2ydy \right )</math>

@@ Line 71: / Line 71: @@
 a.)
 <math>W = \left ( \int_0^Q \vec F \cdot d \vec r + \int_Q^P \vec F \cdot d \vec r\right )</math>
-:<math>= \left ( \int_0^1 F(x,0) dx + \int_0^1  F(1,0)dy \right )</math>
+:<math>= \left ( \int_0^1 \vec F \cdot dx \hat i + \int_0^1  \vec F \cdot dy \hat j\right )</math>
 :<math>= \left ( \int_0^1 x^2 dx + \int_0^1  2ydy \right )</math>
 :<math>= \left ( \frac{1}{3} + 1 \right ) = \frac{4}{3} </math>

← Older revision		Revision as of 15:16, 22 September 2014
Line 68:		Line 68:
	<math>W = \left ( \int_0^Q \vec F \cdot d \vec r + \int_Q^P \vec F \cdot d \vec r\right )</math>		<math>W = \left ( \int_0^Q \vec F \cdot d \vec r + \int_Q^P \vec F \cdot d \vec r\right )</math>
	:<math>= \left ( \int_0^1 F(x,0) dx + \int_0^1 F(1,0)dy \right )</math>		:<math>= \left ( \int_0^1 F(x,0) dx + \int_0^1 F(1,0)dy \right )</math>
		+	:<math>= \left ( \int_0^1 x^2 dx + \int_0^1 2ydy \right )</math>
		+	:<math>= \left ( \frac{1}{3} + 1 \right ) = \frac{4}{3} </math>

@@ Line 65: / Line 65: @@
 ===Problem 4.2===
-a.) W = \left ( \int_0^Q \vec F \cdot d \vec r + \int_Q^P \vec F \cdot d \vec r\right )
+a.)
 [[Forest_UCM_Energy#KE_.26_Work]]

← Older revision		Revision as of 03:06, 22 September 2014
Line 56:		Line 56:

	: <math>T_2-T_1 = \int_1^2 \sum_{i=1}^n \vec {F}_i \cdot d\vec r</math>		: <math>T_2-T_1 = \int_1^2 \sum_{i=1}^n \vec {F}_i \cdot d\vec r</math>
		+
		+	You can either do the summation first and then integrate or integrate and then do the sum
		+
		+	In practice you usually do calcualte the integral part first by calculating the work done by each force then add them all up.
		+

@@ Line 37: / Line 37: @@
 : <math>T_2-T_1 = \sum \vec F_{\mbox{net}}  \cdot d\vec r = \int_1^2 \vec F_{\mbox{net}}  \cdot d\vec r</math> in the limit of dr approaching zero
 ;Negative Work?
@@ Line 48: / Line 45: @@
 And the kinetic energy decreases
 ===Problem 4.2===

← Older revision		Revision as of 02:50, 22 September 2014
Line 37:		Line 37:
	: <math>T_2-T_1 = \sum \vec F_{\mbox{net}} \cdot d\vec r = \int_1^2 \vec F_{\mbox{net}} \cdot d\vec r</math> in the limit of dr approaching zero		: <math>T_2-T_1 = \sum \vec F_{\mbox{net}} \cdot d\vec r = \int_1^2 \vec F_{\mbox{net}} \cdot d\vec r</math> in the limit of dr approaching zero

		+	remember
		+
		+	:<math>\vec F_{\mbox{net}} = \sum_{i=1}^n \vec {F}_i</math>

	;Negative Work?		;Negative Work?