Forest UCM NLM BlockOnIncline - Revision history

Foretony: /* Step 5: Used Newton's second law */

2014-09-08T17:48:25Z

Step 5: Used Newton's second law

2014-08-24T14:08:34Z

Step 5: Used Newton's second law

2014-08-24T14:03:38Z

Step 5: Used Newton's second law

2014-08-24T13:58:25Z

the problem

2014-08-24T13:55:58Z

the problem

2014-08-24T13:53:07Z

Step 5: Used Newton's second law

2014-08-24T13:52:46Z

Step 5: Used Newton's second law

2014-08-24T13:51:56Z

Step 5: Used Newton's second law

2014-08-24T13:51:24Z

Step 5: Used Newton's second law

2014-08-24T13:49:06Z

Step 5: Used Newton's second law

← Older revision		Revision as of 17:48, 8 September 2014
Line 47:		Line 47:
	: <math>\int dt = \int \frac{dv}{k \left ( v_t^2 - v^2 \right)}</math>		: <math>\int dt = \int \frac{dv}{k \left ( v_t^2 - v^2 \right)}</math>

		+
		+	before we had <math>(v dv)</math> in the numerator and could do a <math>u du</math> substitution, but not this time, use an integral table.

	Integral table <math>\Rightarrow</math>		Integral table <math>\Rightarrow</math>

@@ Line 45: / Line 45: @@
-: <math>\int_0^t dt = \int_{v_i}^v \frac{dv}{k \left ( v_t^2 - v^2 \right)}</math>
+: <math>\int dt = \int \frac{dv}{k \left ( v_t^2 - v^2 \right)}</math>
@@ Line 71: / Line 71: @@
 substituting
-:<math>\int_0^t dt = \int_{v_i}^v \frac{dv}{k \left ( v_t^2 - v^2 \right)}</math>
+:<math>\int dt = \int \frac{dv}{k \left ( v_t^2 - v^2 \right)}</math>
 :<math>t = \frac{-i}{kv_t} \tan^{-1} \left ( \frac{iv}{v_t}\right )   = \frac{1}{kv_t} \tanh^{-1} \left ( \frac{v}{v_t} \right ) </math>

@@ Line 38: / Line 38: @@
 :<math> \Rightarrow v_t^2 = \frac{g \sin \theta}{k}</math>
-Insert the terminal velociy constant into Newton's second law
+;This means that the block does not stop sliding but instead it reaches a minimal (terminal) velocity.
 :<math>\sum F_{ext} = mk \left ( v_t^2 - v^2 \right)  = ma_x = m \frac{dv_x}{dt}</math>

← Older revision		Revision as of 13:58, 24 August 2014
Line 6:		Line 6:


−	~~What is its~~ fall ~~time if it falls~~ a distance <math>x</math>?	+	How long does it take to fall a distance <math>x</math>?

	[[File:TF_UCM_InclinedPlaneWfriction.png \| 200 px]]		[[File:TF_UCM_InclinedPlaneWfriction.png \| 200 px]]

@@ Line 1: / Line 1: @@
 =the problem=
-Consider a block of mass m sliding down the inclined plane shown below with a frictional force that is given by
+Consider a block of mass <math>m</math> is sliding down the inclined plane shown below with a frictional force that is given by
 :<math>F_f = kmv^2</math>
 [[File:TF_UCM_InclinedPlaneWfriction.png | 200 px]]
 =Step 1:  Identify the system=

@@ Line 89: / Line 89: @@
 solving for the fall time
-:<math>t = \frac{\cosh^{-1} \left ( e^{kx} \right)}{kv_t}= \frac{\cosh^{-1} \left ( e^{kx} \right)}{\sqrt{kg \sint \theta}}</math>
+:<math>t = \frac{\cosh^{-1} \left ( e^{kx} \right)}{kv_t}= \frac{\cosh^{-1} \left ( e^{kx} \right)}{\sqrt{kg \sin \theta}}</math>
 [[Forest_UCM_NLM#Block_on_incline_with_friction]]

← Older revision		Revision as of 13:51, 24 August 2014
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	:<math>x = \frac{1}{k} \ln \left ( \cosh (k v_t t) \right )</math>		:<math>x = \frac{1}{k} \ln \left ( \cosh (k v_t t) \right )</math>
		+
		+	solving for the fall time
		+
		+	:<math>t = \frac{\cosh^{-1} \left ( e^{kx} \right}}{kv_t}</math>
		+

	[[Forest_UCM_NLM#Block_on_incline_with_friction]]		[[Forest_UCM_NLM#Block_on_incline_with_friction]]

← Older revision		Revision as of 13:49, 24 August 2014
Line 84:		Line 84:
	::<math>\int \tanh (x) dx = \ln \left ( \cosh (x) \right )</math>		::<math>\int \tanh (x) dx = \ln \left ( \cosh (x) \right )</math>

		+
		+	:<math>x = \frac{1}{k} \ln \left ( \cosh (k v_t t) \right )</math>

	[[Forest_UCM_NLM#Block_on_incline_with_friction]]		[[Forest_UCM_NLM#Block_on_incline_with_friction]]