The Energy Principle - Revision history

Karen: /* Mathematical Models */

2024-12-01T04:45:01Z

Mathematical Models

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	<b> Thermal Energy </b><br>		<b> Thermal Energy </b><br>

			Thermal energy is the energy associated with the temperature of the system, and is the internal energy of a system due to the random motion of its particles.

	The formula for the thermal energy change is:		The formula for the thermal energy change is:

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	where:		where:
	* <math> \Delta E_{thermal} </math> is the thermal energy or heat transfer (in joules, J),		* <math> \Delta E_{thermal} </math> is the thermal energy or heat transfer (in joules, J),
	* ''m'' is the mass of the substance (in ~~kilograms~~, kg),		* ''m'' is the mass of the substance (in grams, g),
	* ''c'' is the specific heat capacity of the substance (in joules per kilogram per degree Celsius, J/(kg·°C)),		* ''c'' is the specific heat capacity of the substance (in joules per kilogram per degree Celsius, J/(kg·°C)),
	* ''ΔT'' is the change in temperature (in degrees Celsius or Kelvin, °C or K).		* ''ΔT'' is the change in temperature (in degrees Celsius or Kelvin, °C or K).

2024-12-01T04:41:57Z

Mathematical Models

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	<b> Kinetic Energy </b><br>		<b> Kinetic Energy </b><br>

			Translational kinetic energy is the energy of motion for an object moving in a straight line or along a trajectory.

	The formula for basic kinetic energy (translational kinetic energy) is:		The formula for basic kinetic energy (translational kinetic energy) is:

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	* ''m'' is the mass of the object (in kilograms, kg),		* ''m'' is the mass of the object (in kilograms, kg),
	* ''v'' is the velocity of the object (in meters per second, m/s).		* ''v'' is the velocity of the object (in meters per second, m/s).

			Rotational kinetic energy is the energy of motion for a rotating object. We use this when an object is rotating. We can also combine rotational kinetic energy with translational kinetic energy when the object both translates and rotates.

	The formula for rotational kinetic energy is:		The formula for rotational kinetic energy is:

2024-12-01T04:39:57Z

Mathematical Models

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	<b> Rest Mass Energy </b><br>		<b> Rest Mass Energy </b><br>
			Rest mass energy is the energy inherent in an object due to its mass, even when it is not in motion. This energy will only change if the system we are studying subatomic particles, nuclear reactions like nuclear fission or fusion, or if we are dealing with relativistic systems.

	The formula for rest mass energy, derived from Einstein's theory of relativity, is:		The formula for rest mass energy, derived from Einstein's theory of relativity, is:

2024-12-01T04:19:02Z

Difficult

← Older revision		Revision as of 00:19, 1 December 2024
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	<math> v_f = \sqrt{\frac{ks_i^2}{m} + v_i^2} = \sqrt{\frac{(20)(-2.5)^2}{8} + 3^2} </math>		<math> v_f = \sqrt{\frac{ks_i^2}{m} + v_i^2} = \sqrt{\frac{(20)(-2.5)^2}{8} + 3^2} </math>

	<math> v_f = 14.9 m/s </math>		<math> v_f = 4.96 m/s </math>

	==Connectedness==		==Connectedness==

2024-12-01T04:18:36Z

Difficult

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	Solve for <math> v_f </math>		Solve for <math> v_f </math>

	<math> v_f = \sqrt{\frac{ks_i^2}{m} + v_i^2} = \sqrt{(2)(9.8~~)(10)~~ + 5^2} </math>		<math> v_f = \sqrt{\frac{ks_i^2}{m} + v_i^2} = \sqrt{\frac{(20)(-2.5)^2}{8} + 3^2} </math>

	<math> v_f = 14.9 m/s </math>		<math> v_f = 14.9 m/s </math>

2024-12-01T04:15:00Z

Difficult

← Older revision		Revision as of 00:15, 1 December 2024
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	Solve for <math> v_f </math>		Solve for <math> v_f </math>

			<math> v_f = \sqrt{\frac{ks_i^2}{m} + v_i^2} = \sqrt{(2)(9.8)(10) + 5^2} </math>

			<math> v_f = 14.9 m/s </math>

	==Connectedness==		==Connectedness==

2024-12-01T04:13:38Z

Difficult

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	* <math> s_f = 0 </math>		* <math> s_f = 0 </math>

	'''Step ~~One~~: Apply the Energy Principle'''		'''Step Two: Apply the Energy Principle'''

	<math> \Delta E = W + Q </math>		<math> \Delta E = W + Q </math>

2024-12-01T04:12:59Z

Difficult

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	'''Step One: Apply the Energy Principle'''		'''Step One: Apply the Energy Principle'''
			<math> \Delta E = W + Q </math>

			<math> \Delta K + \Delta U_s = 0 </math>

			<math> \frac{1}{2} m (v_f^2 - v_i^2) + \frac{1}{2} k (s_f^2 - s_i^2) = 0 </math>

			since <math> s_f </math> = 0 we can get rid of <math> s_f^2 </math>

			<math> \frac{1}{2} m (v_f^2 - v_i^2) = \frac{1}{2} ks_i^2 </math>

			We can cancel out the <math> \frac{1}{2} </math> on both sides of the equation

			<math> m (v_f^2 - v_i^2) = ks_i^2 </math>

			Solve for <math> v_f </math>

	==Connectedness==		==Connectedness==

2024-12-01T04:08:30Z

Difficult

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 ===Difficult===
-A horizontal spring with stiffness k = 20 N/m and relaxed length <math> L_0 </math> = 5 m is fixed to a wall and attached to a block of mass m = 8 kg on the other end (no friction between the block and the table). Right now, the spring is compressed to a length of L = 2.5 m, and the block moves ot the left with an initial speed of 3 m/s.
+A horizontal spring with stiffness k = 20 N/m and relaxed length <math> L_0 </math> = 5 m is fixed to a wall and attached to a block of mass m = 8 kg on the other end (no friction between the block and the table). Right now, the spring is compressed to a length of L = 2.5 m, and the block moves to the left with an initial speed of 3 m/s.
 How fast will the block move when the spring is relaxed?
 ==Connectedness==

2024-12-01T03:59:24Z

Difficult

← Older revision		Revision as of 23:59, 30 November 2024
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	===Difficult===		===Difficult===

			A horizontal spring with stiffness k = 20 N/m and relaxed length <math> L_0 </math> = 5 m is fixed to a wall and attached to a block of mass m = 8 kg on the other end (no friction between the block and the table). Right now, the spring is compressed to a length of L = 2.5 m, and the block moves ot the left with an initial speed of 3 m/s.

			How fast will the block move when the spring is relaxed?

	==Connectedness==		==Connectedness==