# Work: Example 2

by Jared Rovny

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00:01 Let's see exactly how this works in an example.

00:04 If a block has mass m, it's sliding smoothly along so no friction to velocity v, and then it encounters a slope, and this slope has an angle of theta and a coefficient of kinetic friction, mu-sub-k.

00:16 So in other words, the slope has friction, the sliding motion, there is no friction.

00:21 And so the question is, how high up the slope does it go.

00:24 So first, before you try this or I do it as well, notice that what we're doing here is now talking about things just in terms of variables, m's and thetas and v's.

00:33 So we're not gonna actually have any numbers in this problem.

00:36 The reason for that is that we need to get more and more into the habit of being able to solve problems purely in terms of the variables, because then if we solve it in terms of the variables, not only can we understand something about the dynamics in how things are working but we can also have a better and more robust way to find the answers without plugging in numbers too early and possibly getting into a mess.

00:57 So, first give this problem a try on your own, see if you can find how high up the slope this object would go and if you've done that, we're going to do it ourselves as well here.

01:06 So first of all, hopefully your analysis of the problem looks something like this.

01:11 You have an object moving with a velocity v, it meets a slope, goes up the slope with gravity pulling it downwards the entire time.

01:18 It also had the force of friction which, remember, always opposes motion.

01:22 So the motion of the object is up the slope which means friction will fight that motion up the slope, and there's also a component of gravity down the slope, Fg times the sine of theta.

01:31 As usual, we have our component of gravity also into the slope, Fg times the cosine of theta, and the normal force keeping the object from falling into the slope.

01:42 So with this picture in mind, let's see if we can figure out how much work was done so that we can use work to find how high up the slope the object would go.

01:51 So initially, our object has some kinetic energy.

01:54 It's moving along and it has a kinetic energy of 1/2 mass times velocity squared.

01:59 The important idea in this problem is that once the object goes up the slope, it slides up a certain distance and then it stops.

02:06 So it stops moving at some height.

02:07 And once it stops moving, it has no more velocity, so the final kinetic energy is zero.

02:12 So just knowing the kinetic energy is we actually know how much work was done on our object.

02:17 So we can say, the change in kinetic energy is equal to the final kinetic energy, zero, minus the initial kinetic energy.

02:26 In other words, our object lost 1/2mv squared, that is change in kinetic energy.

02:33 From our work energy theorem, this will be equal to the work.

02:36 So now let's see if we can write an expression for the work done on our object and hopefully we will be able to find the distance that the object slid up the slope from this expression for work, since work has forces and distances in the equation.

02:48 So let's see what the work looks like.

02:50 We have that the work equals force times the distance.

02:55 So let's analyze all the forces acting on our object.

02:58 So remember, the distance, the direction of displacement of the object is up the slope and it's acting, it's moving upwards during this whole path right here.

03:06 The forces in the direction of the distance are the frictional force as well as this component of the gravitational force which are here and here in the diagram.

03:15 Notice also that these components of the force, the force into the slope, the force out of the slope, are perpendicular to the direction of motion, one is up, one is down.

03:24 The direction of motion is this way, so each of these is a 90 degree angle, remember the cosine of 90 is zero.

03:32 And so for this reason, these forces will not contribute to the work done on our object.

03:38 So that's right, what the total work done is, we have the work done by gravity, which is Fg times the sine of the angle of your slope times the distance that your object moves.

03:50 So I'm gonna use distance for this distance here, this distance is d that it went up the slope, this is what we're trying to solve for.

03:57 And then we also have the force of friction on our object, so we have force of friction on our object and then this is also multiplied by d.

04:07 And now we have to think about this cosine of theta because remember, the distance is up the slope and you have these two forces acting down the slope, which means the angle is 180 degrees. So let's write it like this.

04:20 So the cosine of 180 degrees is minus 1.

04:26 So this means that these two forces here, the gravitational force and the frictional force are contributing to the work expression but their contributing with a negative sign and since they both do that, let's put a negative sign on here, so this is coming from our cosine of theta.

04:42 We have a negative sign from our cosine of theta.

04:44 So, now we have an expression for the total work done.

04:47 It's minus these 2 quantities. So, now let's solve for the distance.

04:52 We know that the work, we have an expression for work here but we also have an expression for work here which we found from the change of kinetic energy.

05:00 So now we're ready to solve and we can equate these two to find the distance that the object travelled.

05:05 So, first we have minus this Fg times the sine of theta plus the force of friction.

05:13 I factored out the distance d, this equals minus 1/2mv squared.

05:20 So, these are just our two expressions for work, one found by the expression for work as force times distance, the other found from the expression for work as the change in kinetic energy.

05:28 Now all we have to do is solve for d. I'll multiply both sides by negative 1 here and then we can say that the distance that the object went is 1/2mv squared and then we divide by the frictional force, Fg sine of theta sorry, the gravitational force and then the frictional force right here.

05:49 Now there's one last step which is to simplify these.

05:52 We know that, so I'll make a little box up here to remind ourselves of a few facts from earlier work.

05:58 We know that the force of gravity is mass times gravity.

06:01 We also knew something about the frictional force.

06:03 If you remember this, the frictional force is the coefficient of kinetic friction times the normal force.

06:09 And what we did in our Newton's law problems is said because there's no motion in or out of the slope, these two forces must be the same, the normal force and the frictional force.

06:20 So, let's clear these out a little bit so we can see.

06:22 The normal force and the frictional force must be the same because otherwise our object will be moving into or out of the slope and so this FN is simply equal to the force into the slope, Fg cosine of theta.

06:34 And I mentioned when we were doing this for the first time with Newton's law problems that this would come up a lot and sure enough here it is, so keep this in mind, this idea that you can always find the normal force, and as I said, we already found Fg to be equal to mg.

06:49 So writing these we now have 1/2mv squared divided by mg times the sine of theta plus and then our frictional force which is mu-sub-k and then the normal which is Fg cosine of theta, mg cosine of theta as we just found. And then we're pretty well done.

07:14 We have a little bit of simplifying we can do, maybe cancel the m's from the top and the bottom by factoring, so this is equal to v squared over 2 where I'm putting the 2 down there from this factor of 1/2, and then we also have g which I can factor from the denominator times the sine of theta plus the cosine of theta and so this, sorry and one last thing which is that we need to remember, mu-sub-k here.

07:42 Always remember that frictional force, so plus mu-sub-k from the friction times the cosine of theta.

07:50 And this is our nice very general result for the distance that the object travelled along the slope d.

07:57 So in a given problem, you will be able to solve in terms of the variables exactly like we've done here and just write it in terms of the variables that you were given in the problem.

08:06 And I always recommend that you plug in your final numbers, the numbers that you're given, way at the very end because often if you plug them in too early, you can lose track of units, you can lose track of how many times you've used your calculator and how many significant figures you might have lost or not lost and also you lose the advantage of getting to simplify like we did here, cancelling out the masses or factoring out a factor of g.

08:27 So, just by doing the calculations one time, at the very end with all of your numbers together, you can save yourself a lot of room for error, so I always recommend that as well.

08:36 So with this example, we have a good overview of how you could use work both as change in kinetic energy as well as force times distance to solve an example of a work problem in finding change in kinetic energy or distances that objects move.

08:50 And this completes our introduction to work as both a force times a distance as well as a change in kinetic energy of objects.

08:57 We'll explore this idea of work a little more in the coming lectures and then move to the idea of momentum and then be finished with our mechanics and ready to tackle some very practical examples of physical laws.

09:09 Thanks for listening.

The lecture Work: Example 2 by Jared Rovny is from the course Work.

### Included Quiz Questions

1. 496J
2. 649J
3. 946J
4. 49.6J
5. 94.6J
1. 4.8 x 1029 J
2. 2.5 x 1029 J
3. 5.6 x 1029 J
4. 3.9 x 1029 J
5. 2.8 x 1029 J

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