WEBVTT physics/ap-physics-1-2/fullerton
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Hi everyone and welcome back to Educator.com.
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In this mini-lesson, we are going to do page 1 of the APlusPhysics worksheet on work and power and you can find that worksheet from the link down below the video.
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With that let us dive right in.
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Problem 1 -- The work done in accelerating an object along a frictionless, horizontal surface is equal to a change in the objects...?
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Well, this is just the Work Energy Theorem.
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The work done is equal to the change in the object's kinetic energy.
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Number 2 -- The graph below represents the relationship between the work done by a student running up a flight of stairs at the time of ascent.
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What does the slope represent?
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Well, slope is rise/run and that is going to be in terms of work and time and work divided by time, as you recall, is power.
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That is going to be the power output of the student.
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Number 3 -- A student does 60 J of work, pushing a 3 kg box up the full length of a ramp that is 5 m long.
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What is the magnitude of the force applied to the box to do this work?
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Well, work is going to be force applied times the distance over which it is applied.
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If we are looking for the force, that is going to be the work divided by the displacement.
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The work done was 60 J. The displacement was 5 m, therefore our force must be 12 N -- Answer Number 3.
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Looking at Number 4 -- We have a boat weighing 900 N and it requires a horizontal force of 600 N to move it across the water at 15 m/s.
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The boats engine must provide energy at the rate of...?
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Well, the rate an energy is provided at, that is power and power can be solved for using force times velocity.
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Our applied force is 600 N.
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Our velocity is 15 m/s, so that is going to be 600 × 15 or 9,000 W, which is 9 × 10³ W -- Answer Number 4.
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On to Number 5 -- A motor used 120 W of power to raise a 15 N object, so the force required was 15 N in a time of 5 s.
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Through what vertical distance was the object raised?
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Well, a couple of different ways we could do that here.
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If power is force times velocity and velocity in this case is just going to be displacement over time, that will be force times our displacement divided by (t).
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We want to solve for the displacement, Δy, so I would say Δy is power times time divided by force...
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...which is going to be 120 W × (time) 5 s/15 N (force) so that is going to be 600/15 or 40 m -- Answer Number 3.
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Number 6 -- The diagram below shows points (A), (B), and (C) at or near Earth's surface.
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As a mass is moved from (A) to (B), 100 J of work are done against gravity.
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What is the amount of work done against gravity as an identical mass is moved from (A) to (C)?
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Well, as we move from (A) to (C) -- remember gravity is a conservative force; it is independent of the path, so the work done against gravity is going to have to be the same thing because (C) is at the same height as (B), so the correct answer is Number 1, 100 J.
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Number 7 -- One watt is equivalent to 1.
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Well, a watt is a unit of power, which is work divided by time and units of work are joules per second, so a joule per second is 1 W.
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Number 8 -- Two weightlifters, one 1 1/2 m tall and one 2 m tall, raised identical 50 kg masses above their heads.
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Compared to the work done by the weightlifter, who is 1.5 m tall, the work done by the weightlifter who is 2 m tall is...?
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Well, that must be greater. The weightlifter who is 2 m tall, lifts the mass further; he does work over a greater distance and in that case you are going to have more work done.
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One last problem here -- A 40 kg student runs up a staircase to a floor that is 5 m higher than her starting point in 7 seconds.
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The students power output is...?
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Well, power is going to be work divided by time, but work is force times displacement divided by time.
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In this case, the force that the student must overcome is the own weight of the student, so that is going to be mass times acceleration due to gravity times displacement divided by time.
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Therefore, power is going to be mass (40 kg), acceleration due to gravity (9.8 m/s²), change in position -- well 5 m in a time of 7 s, so when I put all that together, 40 × 9.8 × 5/7, I get 280 W -- Answer Number 2.
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All right, if those went well -- Terrific -- you are ready to move on to the AP level questions and if it did not go so well, now would be a great time to go back and review the unit, the lesson on work and power.
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Thanks so much for your time everyone and make it a great day.