Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a $60^\circ$ slope at constant speed, as shown in Figure 7.36. The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by friction as the sled moves 30.0 m along the hill? (b) How much work is done by the rope on the sled in this distance? (c) What is the work done by the gravitational force on the sled? (d) What is the total work done?
This is College Physics Answers with Shaun Dychko. This poor ski accident victim is being lowered down this slope at constant speed which means acceleration is zero and it means all the forces are balanced. Now we'll consider this diagram over here that includes all the forces on the person. The only force this diagram is missing is the normal force applied by the surface on the skier. There are four forces, tension, friction, both directed upwards, and number four is perpendicular and gravity straight down. We're going to align our coordinate system parallel and perpendicular to the slope. That'll make our work easier. Because then we don't need to worry about components for three of the forces. Only one of the forces, namely gravity has to be turned into components, a parallel component and the perpendicular component. This angle down here's the same as the angle of the slope theta, which is 60 degrees. Let's talk about why. You could say that imagine this triangle here, this angle up in this top corner here would be 90 minus theta, because this is a right triangle that I've drawn yellow, and so this is 90, this is theta and the total triangle has to add up to 180. This top angle here and this angle theta in the yellow triangle have to add up to 90 which plus this 90 here would make it 180. That makes this part here, 90 minus whatever theta is. Then the next thing I would say is that I would then look at this slope and this dotted line here, they're parallel lines and that makes this angle up here a corresponding angle to this one which was the top angle in this triangle that I had before, and that makes it equal so that makes this bit here between the perpendicular component and gravity makes this 90 minus theta. This is a right triangle here. Let's erase this and talk about this other right triangle here, the gravity force triangle. This is a right triangle, and that makes this angle down here 90 since this angle here and this one here in the green triangle have to add up to 90 which because this one being a right angle, the total angles have to add up to 180. This is 90 minutes theta, and if this plus this has to make 90. We have 90 minus theta equals 90-- Oops, I didn't say that right. If you have 90 minus theta plus this thing that we don't know here, suppose that has to equal 90, then that means this angle that we don't know here is going to be theta, because we will subtract 90 from both sides and then add theta to both sides. You'd be left with the 90 disappearing, and an alpha equals theta. I called this little bit here alpha, but it's theta so I'm labeling it just theta. All right, maybe that's a bit of a long way to explain it. What we've established at this point is that this angle here is equal to the slope angle, and that's important because we're going to use it to resolve this force of gravity and two components that are perpendicular and parallel to the surface. In question A, we're asked how much work does the force of friction do assuming that they descend 30 meters along the slope and willing to figure out what the force of friction is. One thing that we know is that the normal force is pointing in the positive direction, minus the component of gravity that's perpendicular, which is pointing in the negative direction, that equals zero since there's no acceleration. And we can also say that this particular component of gravity is gravity which is mg. Take away that negative thing that just makes things confusing. It's mg times the cosine of theta because this is the adjacent leg of this triangle, adjacent to the angle theta. We use cosine function to get the adjacent leg of a right triangle. We have mgcos theta substituting for perpendicular component of gravity. Then we also know that the force of friction is the coefficient of friction multiplied by the normal force and that means the normal force is the friction force divided by the coefficient of friction when you divide both sides by mu, then switch the sides around and so we can make a substitution of this in place of normal force. Now we have an expression containing the friction force and we'll rearrange that by adding mgcos theta to both sides. Then multiply both sides by the coefficient of friction and you end up with force of friction is mu mgcos theta. Now the work done by friction is the friction force multiplied by the displacement or the competitive component displacement parallel to the friction force. It's going to be friction force times the distance multiplied by cosine of the angle between the friction force and the displacement, which in this case is 180 degrees, because the displacement is down the slope, whereas friction is up the slope, they are antiparallel. We have 180 degrees between them and that's going to be making negative 1. The work done by the friction is negative. We have mumgcos theta substituted in for the friction force, multiply by the distance times the cosine of the angle between the displacement and the friction. The friction coefficient is 0.1 times 90 kilograms mass of this gear, times 9.8 newtons per kilogram, times cosine of 60 because that's the slope angle. We're getting the component of gravity that's perpendicular to the slope times 30 meters. that they descend along the slope times cos 180 which is negative 1320 Joules. Then in part B, we're asked to figure out how much work is done by the rope on the sled. That's the tension force in other words. The force of friction plus the tension foce, which are both in the positive direction in this picture up here, minus the component of gravity that's parallel to the slope is going to be zero. I mean, it would equal mass times acceleration, but since the person is being descended at constant speed, that means acceleration is zero. So in other words, this turns out zero. We know that the component of gravity that's parallel to the slope is the force of gravity multiplied by sine of theta because this is now the opposite leg that we're looking for in this gravity triangle, so opposite to theta. We use sine function to get the opposite leg. We have mg sine theta as the component of gravity parallel to the slope. We substitute that in for F g parallel. Then we're going to solve this for F T by adding mg, sine theta to both sides and then subtracting the force of friction from both sides. The tension force is mg sine theta minus force of friction. Force of friction, we have an expression for that up here, which is mumgcos theta. We substitute that in mumgcos theta here in place of force of friction, and then factor out the common factors mg. We have mg times bracket sine theta minus mucos theta. Now the work done by the tension force is the tension force times the displacement times cosine of the angle etween the displacement and the tension force, and that's 180 degrees again, just as it was above for the force of friction. We substitute in our expression for the tension force. Then we have 90 kilograms times 9.8 newtons per kilogram times sine 60 minus 0.1 times cos 60 times 30 meters times negative one, which is negative 2.16 times 10 to the 4 Joules. That's the work done by the tension force. Then we have the work done by gravity, and so we find the component of gravity that's parallel to the displacement, that's mg sine theta. Then we multiply by that distance times the cosine of the angle between the parallel component and displacement but that's going to be zero degrees because they're both in the same direction, parallel component, gravity's down the slope and so is the displacement. Then we plug in our numbers. We have 90 kilograms times 9.8 newtons per kilogram times sine 60 times 30 meters times cos of zero, which is 2.29 times 10 to the 4 Joules. Then lastly in part D, we find the total work done. That's gonna be confusing any attention working and total work, let's call that the net work done. The net work done is the work done by friction plus the work done by the tension force plus the work done by gravity. We have negative 1,323 Joules work done by friction, minus 21,592 Joules, work done by the tension force, plus 22,915 Joules, work done by gravity, and this works out to zero. We expected this because the total work done or the net work done by a force, or the net work done on an object by all the forces equals the change and the kinetic energy, but since this person is moving at constant speed, there's no change in their kinetic energy, their velocity is constant. This network should be zero as expected and so this checks out with what we were expecting.
Submitted by Richarkc on Sat, 03/21/2020 - 17:15
I watch a lot of your videos and find them very helpful but your explanation for part A had me so completely lost. I also think the drawing was too small for you to add in the extra triangles. It was hard to see what you were doing or circling.
Submitted by ShaunDychko on Sat, 03/21/2020 - 17:58
OK, thanks for this good constructive feedback. I've flagged the video for a re-do. It won't happen soon since I'm still finishing the even numbered problems for other chapters, but I'll get around to it eventually.
Submitted by valerie_508 on Wed, 09/30/2020 - 19:55
How would one find the tension of the rope in this scenario?