Power - Revision history

Lheiman07: /* Difficult */

2026-10-30T05:22:57Z

Difficult

← Older revision		Revision as of 00:22, 30 October 2026
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	A 1200 kg elevator is lifted vertically from rest through a height of 45 m in 15 seconds. At the end of the lift, the elevator is moving upward at a speed of 6 m/s. Assuming the motor operates at a constant power output, calculate the average power output required to lift the elevator. Ignore friction and air resistance.		A 1200 kg elevator is lifted vertically from rest through a height of 45 m in 15 seconds. At the end of the lift, the elevator is moving upward at a speed of 6 m/s. Assuming the motor operates at a constant power output, calculate the average power output required to lift the elevator. Ignore friction and air resistance.

	~~‘’‘Step~~ 1:~~’’’~~ Identify the work done by the motor.		'''Step 1:''' Identify the work done by the motor.

	The motor must provide enough energy to increase both the gravitational potential energy and the kinetic energy of the elevator.		The motor must provide enough energy to increase both the gravitational potential energy and the kinetic energy of the elevator.

	W = \Delta U_g + \Delta K		:<math>W = \Delta U_g + \Delta K</math>

	~~‘’‘Step~~ 2:~~’’’~~ Calculate the change in gravitational potential energy.		'''Step 2:''' Calculate the change in gravitational potential energy.

	\Delta U_g = mg\Delta y		:<math>\Delta U_g = mg\Delta y</math>

	= (1200 kg)(9.8 m/s^2)(45 m)		:<math>= (1200 \ kg)(9.8 \ m/s^2)(45 \ m)</math>

	= 529200 J		:<math>= 529200 \ J</math>

	~~‘’‘Step~~ 3:~~’’’~~ Calculate the change in kinetic energy.		'''Step 3:''' Calculate the change in kinetic energy.

	Since the elevator starts from rest, its initial velocity is zero.		Since the elevator starts from rest, its initial velocity is zero.

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

	= \frac{1}{2}(1200 kg)(6^2-0^2)		:<math>= \frac{1}{2}(1200 \ kg)(6^2-0^2)</math>

	= 21600 J		:<math>= 21600 \ J</math>

	~~‘’‘Step~~ 4:~~’’’~~ Calculate the total work performed by the motor.		'''Step 4:''' Calculate the total work performed by the motor.

	W = \Delta U_g + \Delta K		:<math>W = \Delta U_g + \Delta K</math>

	= 529200 J + 21600 J		:<math>= 529200 \ J + 21600 \ J</math>

	= 550800 J		:<math>= 550800 \ J</math>

	~~‘’‘Step~~ 5:~~’’’~~ Calculate the average power output.		'''Step 5:''' Calculate the average power output.

	Using the relationship between work, time, and power:		Using the relationship between work, time, and power:

	power = \frac{W}{\Delta t}		:<math>power = \frac{W}{\Delta t}</math>

	= \frac{550800 J}{15 s}		:<math>= \frac{550800 \ J}{15 \ s}</math>

	= 36720 watts		:<math>= 36720 \ watts</math>

	~~‘’‘Final~~ Answer:~~’’’~~		'''Final Answer:'''

	\boxed{power = ~~36,720~~ \~~text{~~ W} = 36.72 \~~text{~~ kW}}		:<math>\boxed{power = 36720 \ W = 36.72 \ kW}</math>

	==Connectedness==		==Connectedness==

Lheiman07 at 05:22, 30 October 2026

2026-10-30T05:22:09Z

← Older revision		Revision as of 00:22, 30 October 2026
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	===Difficult===		===Difficult===

			A 1200 kg elevator is lifted vertically from rest through a height of 45 m in 15 seconds. At the end of the lift, the elevator is moving upward at a speed of 6 m/s. Assuming the motor operates at a constant power output, calculate the average power output required to lift the elevator. Ignore friction and air resistance.

			‘’‘Step 1:’’’ Identify the work done by the motor.

			The motor must provide enough energy to increase both the gravitational potential energy and the kinetic energy of the elevator.

			W = \Delta U_g + \Delta K

			‘’‘Step 2:’’’ Calculate the change in gravitational potential energy.

			\Delta U_g = mg\Delta y

			= (1200 kg)(9.8 m/s^2)(45 m)

			= 529200 J

			‘’‘Step 3:’’’ Calculate the change in kinetic energy.

			Since the elevator starts from rest, its initial velocity is zero.

			\Delta K = \frac{1}{2}m(v_f^2-v_i^2)

			= \frac{1}{2}(1200 kg)(6^2-0^2)

			= 21600 J

			‘’‘Step 4:’’’ Calculate the total work performed by the motor.

			W = \Delta U_g + \Delta K

			= 529200 J + 21600 J

			= 550800 J

			‘’‘Step 5:’’’ Calculate the average power output.

			Using the relationship between work, time, and power:

			power = \frac{W}{\Delta t}

			= \frac{550800 J}{15 s}

			= 36720 watts

			‘’‘Final Answer:’’’

			\boxed{power = 36,720 \text{ W} = 36.72 \text{ kW}}

	==Connectedness==		==Connectedness==

Lheiman07: /* A Mathematical Model */

2026-10-30T03:08:29Z

A Mathematical Model

← Older revision		Revision as of 22:08, 29 October 2026
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	:<math>power = \frac{F \Delta r}{\Delta t} = \frac{W}{\Delta t}</math>		:<math>power = \frac{F \Delta r}{\Delta t} = \frac{W}{\Delta t}</math>

	where '''F''' is force, '''Δr''' is displacement, '''Δt''' is the duration of time and '''W''' is work.		where '''F''' is force, '''Δr''' is displacement, '''Δt''' is the duration of time and '''W''' is work.

			Power can also be defined as the rate of change of energy transfer:

			:<math>P = \frac{\Delta E}{\Delta t}</math>

			When there is no heat transfer (Q = 0), the change in energy is equal to the work done. Therefore,

			:<math>P = \frac{\Delta E}{\Delta t} = \frac{W}{\Delta t}</math>

			Since work is the dot product of force and displacement, we can further expand the equation:

			:<math>P = \frac{\vec{F}\cdot\Delta\vec{r}}{\Delta t}</math>

			Rearranging gives:

			:<math>P = \vec{F}\cdot\frac{\Delta\vec{r}}{\Delta t}</math>

	It then follows that instantaneous power is		It then follows that instantaneous power is

	:<math>power = F\cdot v</math>		:<math>power = \vec{F}\cdot\vec{v}</math>

	where '''v''' is velocity.		where '''v''' is velocity.

	The SI unit for power is watts (J/s)		The SI unit for power is watts (J/s). The units can also be expressed in terms of fundamental SI units:

			:<math>Watt = \frac{J}{s} = \frac{N\cdot m}{s} = \frac{kg\cdot m^2}{s^3}</math>

			Note that '''W''' can represent either watts or work, depending on the context.

	===A Computational Model===		===A Computational Model===

Aeze3: /* External links */

2015-12-02T14:52:23Z

External links

← Older revision		Revision as of 09:52, 2 December 2015
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	Internet resources on this topic		Internet resources on this topic

			"General Mechanics/Work and Power." - Wikibooks, Open Books for an Open World. Web. [https://en.wikibooks.org/wiki/General_Mechanics/Work_and_Power General Mechanics/Work and Power]

	==References==		==References==

Aeze3: /* A Mathematical Model */

2015-12-02T14:37:28Z

A Mathematical Model

← Older revision		Revision as of 09:37, 2 December 2015
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	When a force is applied over a distance in a unit of time, power is calculated by		When a force is applied over a distance in a unit of time, power is calculated by

	[[File:power(1).jpg]]		[[File:power(1).jpg \|border\|right]]

	:<math>power = \frac{F \Delta r}{\Delta t} = \frac{W}{\Delta t}</math>		:<math>power = \frac{F \Delta r}{\Delta t} = \frac{W}{\Delta t}</math>

Aeze3: /* The Main Idea */

2015-12-02T14:36:49Z

The Main Idea

← Older revision		Revision as of 09:36, 2 December 2015
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	When a force is applied over a distance in a unit of time, power is calculated by		When a force is applied over a distance in a unit of time, power is calculated by

	[[File:power(1).jpg ~~\|border\|right~~]]		[[File:power(1).jpg]]

	:<math>power = \frac{F \Delta r}{\Delta t} = \frac{W}{\Delta t}</math>		:<math>power = \frac{F \Delta r}{\Delta t} = \frac{W}{\Delta t}</math>

Aeze3: /* The Main Idea */

2015-12-02T14:36:30Z

The Main Idea

← Older revision		Revision as of 09:36, 2 December 2015
Line 10:		Line 10:

	When a force is applied over a distance in a unit of time, power is calculated by		When a force is applied over a distance in a unit of time, power is calculated by

			[[File:power(1).jpg \|border\|right]]

	:<math>power = \frac{F \Delta r}{\Delta t} = \frac{W}{\Delta t}</math>		:<math>power = \frac{F \Delta r}{\Delta t} = \frac{W}{\Delta t}</math>
	where '''F''' is force, '''Δr''' is displacement, '''Δt''' is the duration of time and '''W''' is work.		where '''F''' is force, '''Δr''' is displacement, '''Δt''' is the duration of time and '''W''' is work.

	~~[[File:power(1).jpg \|border\|right]]~~

	It then follows that instantaneous power is		It then follows that instantaneous power is

Aeze3: /* The Main Idea */

2015-12-02T14:36:06Z

The Main Idea

← Older revision		Revision as of 09:36, 2 December 2015
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	where '''F''' is force, '''Δr''' is displacement, '''Δt''' is the duration of time and '''W''' is work.		where '''F''' is force, '''Δr''' is displacement, '''Δt''' is the duration of time and '''W''' is work.

	[[File:~~http://www.differencebetween.info/sites/default/files/images/2/~~power(1).jpg \|border\|right]]		[[File:power(1).jpg \|border\|right]]

	It then follows that instantaneous power is		It then follows that instantaneous power is

Aeze3: /* Simple */

2015-12-02T14:26:58Z

Simple

← Older revision		Revision as of 09:26, 2 December 2015
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	A certain motor is capable of doing 3000 J of work in 12 s		A certain motor is capable of doing 3000 J of work in 12 s
	What is the power output of this motor?		What is the power output of this motor?


	:<math>power = \frac{W}{\Delta t} = \frac{3000 J}{12 s} = 250 watts </math>		:<math>power = \frac{W}{\Delta t} = \frac{3000 J}{12 s} = 250 watts </math>

Aeze3: /* Simple */

2015-12-02T14:26:45Z

Simple

← Older revision		Revision as of 09:26, 2 December 2015
Line 35:		Line 35:
	A certain motor is capable of doing 3000 J of work in 12 s		A certain motor is capable of doing 3000 J of work in 12 s
	What is the power output of this motor?		What is the power output of this motor?


	:<math>power = \frac{W}{\Delta t} = \frac{3000 J}{12 s} = 250 watts </math>		:<math>power = \frac{W}{\Delta t} = \frac{3000 J}{12 s} = 250 watts </math>