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Work & Power

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Question 1 0:13
  • Question 2 0:29
  • Question 3 0:55
  • Question 4 1:36
  • Question 5 2:18
  • Question 6 3:22
  • Question 7 4:01
  • Question 8 4:18
  • Question 9 4:49

Transcription: Work & Power

Hi everyone and welcome back to

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.0003

With that let us dive right in.0011

Problem 1 -- The work done in accelerating an object along a frictionless, horizontal surface is equal to a change in the objects...?0013

Well, this is just the Work Energy Theorem.0020

The work done is equal to the change in the object's kinetic energy.0022

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.0029

What does the slope represent?0035

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.0038

That is going to be the power output of the student.0049

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.0055

What is the magnitude of the force applied to the box to do this work?0062

Well, work is going to be force applied times the distance over which it is applied.0066

If we are looking for the force, that is going to be the work divided by the displacement.0073

The work done was 60 J. The displacement was 5 m, therefore our force must be 12 N -- Answer Number 3.0081

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.0096

The boats engine must provide energy at the rate of...?0105

Well, the rate an energy is provided at, that is power and power can be solved for using force times velocity.0108

Our applied force is 600 N.0117

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 4.0120

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.0138

Through what vertical distance was the object raised?0152

Well, a couple of different ways we could do that here.0157

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).0160

We want to solve for the displacement, δy, so I would say δy is power times time divided by force...0175

...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 3.0183

Number 6 -- The diagram below shows points (A), (B), and (C) at or near Earth's surface.0202

As a mass is moved from (A) to (B), 100 J of work are done against gravity.0210

What is the amount of work done against gravity as an identical mass is moved from (A) to (C)?0217

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.0222

Number 7 -- One watt is equivalent to 1.0241

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.0245

Number 8 -- Two weightlifters, one 1 1/2 m tall and one 2 m tall, raised identical 50 kg masses above their heads.0258

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...?0266

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.0274

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.0287

The students power output is...?0296

Well, power is going to be work divided by time, but work is force times displacement divided by time.0299

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.0309

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 2.0320

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.0348

Thanks so much for your time everyone and make it a great day.0359