This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

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From the course by University of Houston System

Preparing for the AP Physics 1 Exam

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University of Houston System

35 ratings

This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

From the lesson

Energy and Momentum

Topics include work, energy, conservation of energy, impulse, linear momentum, and conservation of linear momentum. You will watch 3 videos, complete 3 sets of practice problems and take 2 quizzes.Please click on the resources tab for each video to view the corresponding student handout.

- Dr. Paige K. EvansClinical Associate Professor

Math - Mariam ManuelInstructional Assistant Professor

University of Houston

In this module we will discuss the topics of work, energy, and power.

Â The term work is defined in a very specific way in physics.

Â Work is what occurs when a force acts on an object, and

Â it experiences a displacement.

Â To calculate the work done on an object by a constant force, we multiply the object's

Â displacement by the component of the force parallel to that displacement.

Â The equation seen here shows us that the units for

Â work are newtons multiplied by meters, also called a joule.

Â Note that work is a scalar quantity, and does not have direction.

Â We use the word work in our daily lives differently than we would in physics.

Â Let's look at an example where we calculate the amount of work

Â done per person.

Â A businessman carries a briefcase at his side as he

Â walks at a constant speed of 3 meters per second to the left.

Â If the briefcase has a mass of 10 kilograms, and he walks a total

Â distance 50 meters at this speed, how much work does he do on the briefcase?

Â Let's consider the forces acting here.

Â Keep in mind, since the man walks with a constant velocity,

Â no horizontal force is needed to move the briefcase at that constant velocity.

Â So, we have mg, which is the weight of the briefcase directed downwards towards

Â the center of the earth, we have the force applied by the man to carry the briefcase,

Â which is upwards, opposite and

Â equal to mg, and we have the man's displacement directed towards the left.

Â Since the force applied by this man is perpendicular to the displacement,

Â he does no work on the briefcase.

Â >> This problem firsts ask us to calculate work done by each force on this block, and

Â then the total work done on this block.

Â So, taking a look at what forces act on this block to begin with, we have a block

Â that's originally moving to the right with a speed of 4 meters per second.

Â And I have various forces that then act on it.

Â It tells me that there's a tension,

Â acting at an angle of 20 degrees above the horizontal.

Â So this angle here.

Â 20 degrees.

Â It has a mass to it, that means that there is a weight down,

Â mg always acting vertically straight downward.

Â That means there is also a normal force vertically upward.

Â And it also gives me this coefficient of kinetic friction, which means to resist.

Â It's motion to the right.

Â There is a frictional force left parallel to the surface but

Â against the motion of direction.

Â If I want to calculate the work done by each force we have to remember our

Â definition for work.

Â Work is force parallel to a distance.

Â You'll often see that written as force times distance times the cosine of theta.

Â But you have to be careful of that definition,

Â and be very particular about what angle you're using.

Â It's only going to be the cosine of theta if it's the angle between the force and

Â the displacement vectors.

Â If not, what you really need to look at in your diagram is

Â the parallel component of that force in the direction of that displacement.

Â So let's see what I mean by that.

Â Let's find the work done by the normal force.

Â So for the normal force, if we're talking about the work,

Â well the normal force itself is represented by N.

Â Then it travels a distance, d, it tells me it travels to the right here.

Â And I'll draw that on my diagram, so you can see what I mean.

Â It travels to the right.

Â A distance of 10 meters.

Â And that would be my distance d, in this problem.

Â But then I need the cosine of the angle between those two vectors.

Â Notice there is no parallel component.

Â The normal force does not pull it, at all, in the direction of its motion.

Â So my angle, here, between the two,

Â would be 90, if you wanted to use this version of the equation.

Â Then the cosine of 90 is 0, so the work done by normal force is 0.

Â Yes, that force does push on it, but it does no work.

Â It never gives or takes away any energy from the block.

Â Friction, EMG, and tension are all other forces that we can calculate.

Â So, for instance, the weight, force, EMG.

Â The work done by that.

Â Well, it's going to be mg the force,

Â times the distance, times the cosine of the angle between them.

Â Again, the placement is to the right, but mg pulls down.

Â That's a 90 degree angle between those two vectors.

Â Again, 0 Joules.

Â No work done by either of those forces.

Â I can tell in my diagram, though, that tension will do some work, and so

Â will friction.

Â Both of those forces have a component parallel to the component.

Â So let's talk about tension.

Â [SOUND] The tension.

Â Let me show some on this diagram here, so you can see it.

Â The work of the tension would be the force, which is T.

Â But I want the parallel component to the displacement in my diagram.

Â We'll change colors here.

Â It's this component.

Â Not the perpendicular component.

Â I need the parallel component.

Â So that's going to be T cosine of theta because it's the adjacent side of

Â that triangle, times the distance it travels, which is d.

Â And so using the values that they gave me in the problem,

Â the tension was 300 cosine of

Â the angle between those two vectors, which in this problem was 20 degrees.

Â And the distance it traveled which was 10.

Â When I multiply all that out, the work of

Â the tension or, on this block, 2820 joules.

Â Tension, pulled in the direction of the motion, it gave this block energy.

Â What kind of energy?

Â We'll talk about that shortly, but it makes the block speed up.

Â It gives it a kinetic energy.

Â We've got one last force to talk about, which is force of friction.

Â The work done by friction, again,

Â will be our force of friction times the distance times the angle between them, and

Â take a look at what's going on in this diagram up here.

Â The block moves to the right, it's displacement is to the right, but

Â the friction pulls left.

Â That's not a cosine of 0, the angle between those two vectors is 180.

Â They point in opposite directions and the cosine of 180 is a negative 1.

Â So a good rule of thumb, looking up at my diagram, any time a force and

Â a displacement are in opposite directions like this, you have a negative work.

Â This is a scenario where it actually takes energy away.

Â And friction is one of those forces that will always be doing negative work.

Â So my work done by friction, mu times my normal force times my distance.

Â And this is all that's multiplied by negative 1.

Â Now there's something here that we have to keep in mind.

Â The normal force in this scenario is not just mg.

Â This block is being pulled up some by tension, and it's very easy to miss that.

Â That means I need to set up an expression of forces to find the normal

Â force on this block.

Â So, let's do that really quickly.

Â Sum of forces in the y direction equals ma.

Â And I'm setting up in the y direction,

Â because the normal force points in the y direction.

Â It's not accelerating in the y direction, and

Â I have three vertical forces: the normal force, which is straight up on this block,

Â the tension also pulls up, so that's a positive T.

Â I need the vertical component.

Â I need t sine of theta to get me

Â to pull up minus mg equals 0.

Â Solving this for normal force, normal force will equal mg minus t sine theta.

Â In essence the surface has to push back less.

Â Because that tension's already pulling the block up some.

Â It doesn't have to equal the whole mg.

Â It's mg minus this pull upward on the block.

Â That is what I can sub into my equation here, for normal force.

Â So, giving us a little bit more room to do that real quick,.

Â Our equation becomes work of friction, mu m g minus T sine theta, times

Â the distance, and I'm just going to make the negative sign come out front.

Â Plugging in our number, that is a relatively long problem here.

Â Just make sure you're not missing any of your numbers.

Â Your coefficient to connect friction, 0.6.

Â The mass, 30, g, 10.

Â For our problems here, the tension, 300, sin of 20, that angle between them.

Â This gives me that vertical component for the normal force.

Â And it goes a distance of 10 meters.

Â That tells me that the work of friction is negative, like we expected.

Â Friction takes away energy from the box.

Â It was directed in the opposite direction of the displacement.

Â The work of friction, then, is this negative 1184 joules.

Â Compare that to what we had earlier for the work of tension.

Â Those are our two works, so then ask what's the net work on the block.

Â Those are the two numbers that I'm going to use.

Â Just really quickly here, the net work is the sum of works.

Â We have the work done by the tension and a work done by friction.

Â The work done by the tension was 2820.

Â The work done by friction was

Â a negative, this number right up here, 1184.

Â And adding those two together, together gives me the net work.

Â 1636 joules.

Â In the end, more work was done on the box than was taken away.

Â In this case, what I mean by that, is, there's a positive work,

Â which means energy was given to the box.

Â And there was a negative work, that means energy was taken away.

Â In the end more energy was added to the box than taken away, so

Â we will see this box actually speed up, which we'll talk about next.

Â >> So the last time we looked at this question, we solved for

Â the net work done and that was 1,636 newtons.

Â Well, now this question wants us to determine the final velocity after this

Â block has moved ten meters, and

Â then determine the change in velocity, how much did it increase or decrease.

Â So, I'm going to go ahead and use my work energy principle.

Â That principle tells me that the net work done is equivalent to the change

Â in kinetic energy.

Â going to rewrite this, this one-half

Â m v final squared minus one-half m v initial squared.

Â So changing kinetic energy back being final

Â kinetic energy minus initial kinetic energy.

Â I can now plug in values.

Â [SOUND] Like I said, I have the net worth from the last problem.

Â The mass given is 30 kilograms.

Â I don't know what the final velocity is.

Â That's what I'm looking for.

Â My initial velocity is 4 meters per second.

Â And so, now, when I solve for my final velocity,

Â I end up getting the value, 11.18 meters per second.

Â One thing I want to advise you about is that,

Â make sure you remember that the final velocity here is squared.

Â A lot of times students forget that when they're solving for

Â final velocity, they will need to take the square root to get the answer.

Â So, if my final velocity's 11.18, and

Â my initial velocity was 4 meters per second,

Â then my change in velocity is 7.18 meters per second.

Â So, the increase in speed was 7.18 meters per second.

Â >> To calculate the power developed in this block,

Â you need to remember that power is defined as work divided by time.

Â How much work is done per unit time.

Â Well, we want to know the net power developed in this block.

Â So that's the net work.

Â Well that's a number that we actually already calculated.

Â And that's work from all the forces on this block,

Â as we saw a moment ago came out to be about 1,636 joules.

Â We know that number.

Â But notice that I don't know the time.

Â That's one of the problems that we encounter when we're calculating energy.

Â There is no T in those equations to really tell us how long something took.

Â We're going to have to find another way to solve for time.

Â Keep in mind this is the scenario where we

Â have a tension at an angle of 20 degrees above a horizontal,

Â where we have a frictional force resisting that pull to the right.

Â We have a normal force, and we have an mg.

Â None of those forces are changing over time.

Â Since the force is constant,

Â that means that the acceleration of this block is constant.

Â And it's a great opportunity for us to use kinematic equations.

Â So to do that, I'm going to start off with forces.

Â Sum of forces in the x direction on this block,

Â because I know it accelerates in the x direction, and

Â that's what I'm looking for, we're equal to mass of that block times acceleration.

Â So looking at my scenario, right is positive, left is negative.

Â So I'm going to have a t,

Â and now I need the horizontal component of this tension here.

Â That means I need to use the cosine of 20 degrees,

Â the adjacent side of this right triangle.

Â And the vertical component will come up again in the moment when we

Â need need [INAUDIBLE] force.

Â So positive T cosine of 20 minus the frictional force will equal the mass

Â of that block times its acceleration.

Â As we saw earlier, T cosine of 20 remains, but I can sub in with normal

Â force in the coefficient of friction, mu equals mass times acceleration.

Â And remember, the normal force here was not just mg, as we saw earlier.

Â The normal force instead will be, let's see here.

Â Cosine 20 is still there.

Â [SOUND] We're going to have u mg minus the T of the vertical component.

Â This is where it comes up.

Â Because it's not just normal force mg.

Â It's T sine theta.

Â The surface doesn't have to push back as hard,

Â because this tension in the string is pulling up on the block.

Â Will equal ma.

Â Lots of values here, but I'm going to sub in my numbers.

Â The tension was 300, cosine 20 minus the coefficient of friction 0.6.

Â We're talking about a 30 kilogram block, g is 10.

Â Minus tension of 300, sin of 20 degrees again,

Â because the angle between that string and the horizontal,

Â giving us the vertical component, equals the mass 30 times acceleration.

Â Solving for that very carefully, because there's lots of numbers there,

Â I get an acceleration of 5.45 meters per second squared to the right.

Â This block is accelerating in the rightward direction.

Â I can then take this acceleration that's not what I need to know yet.

Â I'm getting close, I need to know time.

Â I'm going to choose this kinematic equation.

Â Since I know the distance as well.

Â Keep in mind this object did not start for rest.

Â It starts with initial velocity of 4 meters per second to the right.

Â So I'm going to have to make sure to include that and

Â not drop off that initial velocity.

Â Other kinetic equations could work.

Â You could use the other two instead of a system of equations to

Â get to the same answer.

Â It told us they moved this block 10 meters.

Â We started with an initial velocity of 4.

Â We don't know the time.

Â We don't know the time.

Â That's what we're looking for.

Â The acceleration we just solved for, and we're solving for, t.

Â This is a quadratic equation.

Â We solved one of these earlier in one of our modules where we showed you how to

Â either use the quadratic equation, or how to graph this in a graphic calculator,

Â which is the preferred method.

Â It's much quicker.

Â You get to use your graphing calculator on the whole AP test.

Â Free response and multiple choice.

Â Solving this for t of n I get t is 1.32 seconds.

Â That's how long this force was acting,

Â this net force acting on the block that causes it to speed up.

Â Coming along then solve for power now.

Â The total work over the time it took.

Â We found earlier that the total work done on the block,

Â 1636 joules, we just found that it took 1.32 seconds.

Â Dividing, I get the total watts,

Â the power developed in this block to be 12 by 39 watts.

Â Be careful that the watts here does not get confused with the variable that we

Â used to represent work.

Â They are different in meaning.

Â Be careful that you don't mix the two up.

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