Practice: You push on 3 kg box against a wall for a distance of 2 m with a 100-N force that makes 53° with the horizontal, as shown. The box-wall coefficient of friction is 0.3. Calculate the work done by:

(a) you,

(b) friction,

(c) gravity.

Subjects

Sections | |||
---|---|---|---|

Intro to Energy | 15 mins | 0 completed | Learn |

Intro to Calculating Work | 36 mins | 0 completed | Learn |

Work By Gravity & Inclined Planes | 40 mins | 0 completed | Learn |

Work By Variable Forces (Springs) | 47 mins | 0 completed | Learn |

Net Work & Kinetic Energy | 39 mins | 0 completed | Learn |

More Work-Energy Problems | 39 mins | 0 completed | Learn |

Power | 20 mins | 0 completed | Learn |

Concept #1: Work Done by Gravity

**Transcript**

Hey guys, so now we're going to talk about the work done by gravity as things are going up and down how much work does gravity do, I think this is pretty straightforward but I want to give you a clean explanation of it so that if you see this in a test you can just knock it out of the park, alright? Let's check it out so the work done by gravity, remember positive work is when a force is helping motions to make you move faster something like that, negative work is when it's hurting motion friction is always pulling against you so it's hurting you motion does negative work and forces that don't affect motion at all do no work or they do zero work, OK? Here's our work equation how we calculate work, now when an object is falling gravity does positive or negative work on it? What do you think based on this definition here helps motion, hurts motion do you think when an object is falling gravity does positive or negative work? Gravity does positive work on it, right? Because it's helping motion its what causes falling in the first place so if you have a mass and it's falling, MG pulls it down so gravity is helping motion they go in the same direction that's what you get, now if an object is going up gravity is pulling against it and gravity does negative work on it it's that simple, OK? Let's talk about one more thing, what about if you're lifting something, well the force to lift something is MG this is a little bit counterintuitive and I'll show you why this might be confusing, so if you're lifting something first of all every object near the earth is going to be pulled down by an MG, OK? To lift it your force to lift is MG, so at first you must look at this and say well if it's 100 down and I pull it with 100 up wouldn't it cancel? And it does you're right it cancels if this is a 100 then your force needs to be a 100.00001 but that's kind of silly so it has to be slightly greater than MG so we just basically set it equals MG, OK? But you understand it needs to be a little more but mathematically kind of rounds to MG so whenever you're lifting something the force the bare minimum force you need to lift it is MG, OK? Now in physics, lifting this is a very specific thing for physics lifting in physics if that's all you get implies that you're taking it from an initial velocity of 0 to a final velocity of 0 so there's a book here I'm going to lift it, not moving.... not moving, right? So, it's either from initial velocity of 0 to a final velocity of 0 and another thing you might hear is that it's you're lifting something with a constant speed, right? So that's the same idea here and the reason why you'll be at constant speed is again if you get the object moving and then these are the same thing so the object accelerates a little bit to start moving but then moves at a constant speed, OK? So, lifting or lifting with a constant speed this could be another one means that the force you need to do that is MG the idea is that you need to barely overcome the force of weight that's pull you down, so what is the work to lift something, right? Well if the force to lift is MG then the work to lift is the force to lift which is MG times the distance, now let's say you're lifting something a height of H, right? So, the distance would be H x the cosine of theta. Now if I pull something up and as a result the object moves up the angle between these two is 0 so this is going to be 0 right here and you know the cosine of 0 is 1 so I'm going to say the work to lift is MGH, now instead of calling this MGH I'm actually going to call this MG delta H because MGH means that implies that you're starting from the ground or maybe you're starting from the ground maybe you're starting from 2 to 3, so your delta H from 2 to 3 is 1 so MG delta H. I hope that you remember MGH, MGH was your potential energy gravitational potential energy is MGH, so MG Delta H is basically your change in gravitational potential energy, OK? change in gravitational potential energy, so long story short work lift is this that's how you're going to calculate it but sort of conceptually it's the change in potential energy and this should make sense if you have a box on the floor it has no potential energy when you lift it gains a 100 Joules of potential energy and that's because you did a 100 Joules of work so you pumped 100 Joules of potential energy into this thing, if you lift the you're doing positive work if you bring it down you're doing negative work because at that point if you move something down you're stealing potential energy from the object, right? You can think about it that way. Gravity does negative work if you lift something because as you're going up gravity is pulling you down so it's going opposite to the displacement so the reason why that's going to happen is because you're going to have MG that's the force of gravity so the work done by MG is MG if you go up that way height but gravity is pulling you down, gravity is pulling you down which means the angle here will be 180, Ok? so this will be the cosine of 180. And this is going to be -1 so I can say that it's -MG Delta H so instead of being the change in potential energy it's going to be the negative change in potential energy which means instead of being (final-initial) it's (initial-final), OK? The important one to remember is this one and this is more of a conceptual one, so notice how these things are completely opposite, OK? So, you can just remember that the work to lift is MG delta H and the work done by gravity as you're moving something up and down is negative MG delta H, OK? If you're moving up then your delta H is positive and if you're moving down then your delta is negative, right? (Final-initial), OK? So, this is pretty straightforward we're going to some examples here and I think you guys are going to get it, alright? That's it MG delta H which is should make sense because it's the gravitational potential energy, you lift something you give it potential energy and that's coming from the work that you did, OK? So, you lift a 3-kilogram box straight up from the floor and place it on a shelf 2 meters tall shelf above the floor so that's let's say there's a 3 kilograms book here or box rather it's on the floor and let's say over here there's a shelf, alright? You're going to move this guy and put it over here. I want to know how much work did you do? And this is very straightforward, OK? The work done by you is the same thing as the work done to lift and the work done to lift is MG delta H, OK? If this is 2 meters high I can see that this is H=0, this is H is 2 so your delta H=+2, OK? Very straightforward, mass is 3, Gravity is 10 this +2 so this is sixty Joules.

For part B. what is the work done by gravity? The work done by gravity or the work done by weight same thing those are used interchangeably technically not the same thing but they're used interchangeably, the work done so it's going to be instead of MGH if you remember it's -MG delta H, now the variables are exactly the same the only thing that happens different here is that you have a negative so it's just going to be -60, alright? Check it out, -3, 10 this is +2 I'm still going up the reason why it's negative it's not because +2 becomes negative but because there's a negative in front of the equation so it's -60 Joules, itÕs says the box then falls to the floor how much work does gravity do then? So, the box is falling here's a 3-kilogram box it's falling so I can say the initial heights is zero...And I'm sorry the initial height is 2 and the final height is 0 so what is my change in height? My change in height is -2, Why? Because I dropped 2, OK? So, if I go here the work done by gravity is always negative MGH or MG delta H, -3 gravity we're using 10 my delta H is -2 right here, OK? Just to be clear this -2 comes from here and if you do this the negative cancel you get a positive 60 joules and this should make sense because gravity does positive work on the way down so you gave this object's 60 Joules of potential energy and then meanwhile was doing negative work against you and when it drops gravity gives it back the 60 Joules of potential energy that it kind of stole on the way up, OK? So, gravity does negative work going down and positive work... I'm sorry negative work going up and positive work going down, OK? That's the gist of it we're going to do a few more problems I want you to try this one it's a little tricky but I think you might be able to get it, letÕs give this one a shot.

Practice: You push on 3 kg box against a wall for a distance of 2 m with a 100-N force that makes 53° with the horizontal, as shown. The box-wall coefficient of friction is 0.3. Calculate the work done by:

(a) you,

(b) friction,

(c) gravity.

Practice: A 70 kg person hikes from the bottom to the top of a 1,000 m hill with varying speeds. The path you take is very irregular, with varying inclinations. How much total work does gravity do on you during the entire hike?

Concept #2: Work on Inclined Planes

**Transcript**

Hey guys in this video I want to talk about work problems, work done by forces. In situations where the object is moving along an inclined plane. And the reason why this requires itÕs own video is because these questions can be pretty tricky and I want you to be ready for some of these things. LetÕs check it out. So first remember theta in the work equation is the angle between the force and the displacement. Right force and displacement they form two vectors, two arrows. ThereÕs a very specific definition as to what theta is. And a lot of questions will give you the wrong angle to try to trick you into using the wrong angle because they want to make sure they know what youÕre doing. So I think itÕs helpful to have sort of a healthy amount of paranoia every time you see a theta to make sure that itÕs the right one. Okay. One quick thing here before we start. When you have an angle, when you have an incline plane like this, the planeÕs length which is this, I can call this L or d, the distance going up or down, is related to itÕs height and to itÕs angle by this equation h equals L sin of theta. I need you to remember this as well as obviously the equation for potential energy m g h and the work done by force which is this one. Okay. IÕm gonna do an example that I want you guys to try one practice problem. So letÕs check this out. I have a 100 kilogram crate sliding at a constant 7, so let me write some of this stuff down. Mass is 100, velocity is 7, itÕs a constant velocity which tells me the acceleration is zero. From the top of a 12 meter long ramp, that makes 37 with the horizontal. So let me draw a little ramp here. And this ramp makes 37 with the horizontal down here. It slides 12 meters so IÕm gonna say that the entire amounts of, the length of your displacement or your distance is 12 and you start up here and youÕre going to slide down. So first we wanna know is what is the crateÕs initial potential energy. So potential initial is potential initial gravitational because thereÕs no spring so the potential here has to be gravitational only, and itÕs just if you remember m g h initial, potential energy m g h. The mass is 100, gravity weÕre going to use 10 just to make it simpler and my original height is I donÕt have it. Right, I donÕt have the original height, but I can find the original height and thatÕs because h is L or d sin of theta so itÕs 12 sin of 37 and 12 sin of 37, I have it here, is 7.2. So thatÕs my height initial and thatÕs whatÕs gonna go right here, 7.2. When I multiply this whole thing I get 7200 Jules as my initial potential energy. Cool. ThatÕs it. The next question says use the work equation, which is this equation right here to calculate the work done by m g x, m g y and friction. So the work done by m g x. Remember the work done by any force, since I want to use this equation, is that force m g x distance cosine of theta. But before I do anything IÕm gonna have to calculate m g x. So letÕs do that real quick. M g x, I hope you remember is m g sin of theta so mass, gravity weÕre gonna use 10 and sin of 37. When you do this real quick you get a 600. M g y is m g cosine of 37 and thatÕs 800. Okay these are forces, this is a height but this is a force so these are Newtons. So m g x is this number right here, okay. So 600 goes right here. What is the distance. Well the distance you moved was 12 and what is the angle. The angle is the tricky one, right. So some people might have thought that that was 37. ItÕs not. ItÕs the whole purpose for the existence of this question is to make the point itÕs not that angle. Okay. Check it out. You are moving this way, and m g x is this way. Those two are parallel to each other which means they make an angle of zero degrees with each other. Okay. So this is the cosine of zero. And when you multiply all of this you get that the work done by m g x is 7200 Jules. I got a positive which should make sense because even though IÕm not going directly down, I am losing height and if you remember, the work done by m g is m g delta h. So if I had an initial energy of 7200 the work done by m g should be 7200 if I fell the whole way down. So that kind of matches up. What about the work done by m g y. So m g y, m g y is 800, the distance is 12 cosine of what? Well youÕre moving this way delta x, m g y is always into the plane so this makes an angle of 90 which means this is zero. And you might remember the work done by m g y is always zero because itÕs always perpendicular to the plane, itÕs always into the plane. The same thing is the work done by normal. So it shouldnÕt be a surprise that we got a zero here. The last point I wanna make is that the work done by mg can be thought of as the work done by m g x plus the work done by m g y, but the work done by m g y will always be zero so the work done by m g is the same as the work done by m g x. Okay. And if you remember, the work done by m g is also m g negative, m g drop in height. In this problem, my change in height was negative so these two negatives cancel and my total work done by m g turned out to be positive which is consistent with the idea that if youÕre going down, m g does positive work. Okay. So thatÕs the basic idea here. Explored a few things. Key points, so we got to use our h equals L sin equation, cool, but a key idea here is that this angle here is zero. Okay, not 37. ThatÕs the angle for the plane. Another point thatÕs important to make is that I do use this angle in these equations here, right. When youÕre decomposing m g x and m g y you have to use the angle at the bottom of the incline, but for work thereÕs a very specific definition which is the angle between the direction or the displacement in the force and in this case m g x and delta x went down. IÕve said that a few times already in this video alone, hopefully it clicks. Let me know if you guys have any questions, and I want you to do this next practice problem. LetÕs give this a shot.

Practice: You push a 2-kg box from the bottom of an inclined plane with 50 N for 10 m. The incline makes 37° with the horizontal, and the box-incline coefficient of friction is 0.6. Find the work done by:

(a) you,

(b) friction, and

(c) gravity.

Example #1: Work on Inclined Planes

**Transcript**

Alright so here's an extra example, this is a little bit weird so I'm going to do this for you guys I have a 5-kilogram box that's pulled at 3 meters per second so mass=5, velocity=3 when I say velocity is 3 I'm implying that the acceleration is 0 which is called equilibrium which also means that if you remember it also means that forces will cancel equilibrium which means the forces cancel, OK? So here is you're being pulled up on a conveyor belt, right? So, this belt is moving there's an object here being moved by the belts, so 5 kilogram objects being pulled up at the velocity of 3 meters/second so there's a force that pulls you up and I'm going to call this F belt, there's a force that pulls you up why is it that there has to be a force? Because otherwise you would be falling due to MGX over here, OK? You're moving with a velocity of 3 for a distance of 10 and the angle here is 37 degrees, OK? A little different how much work is done by gravity? So the work done by gravity is negative MG delta H, mass is negative mass is 5, gravity is 10 and I need the delta H, I don't have delta H but I can find it again, H is L sine of theta, this is a very popular equation for these inclined plane problems to see if you are familiar with how to go from one to the other, the length here is 10 sine of 37 which is simply 6, this 6 goes here and this is -300 joules and it should make sense that it's negative work because as you grow up gravity is doing negative work against you, OK? That's where the number comes from.

Question B is sort of more of a conceptual question which force is pulling the object up? As you go in a conveyor belt like this it pulls you up which force does that, I hope you know this I hope you realize this but if you don't I'll tell you and then hopefully remember F belt is really static friction, right? If I have an object in my arm and I pull my arm and the object comes with it, it's because friction of my arm is dragging along it's causing some traction, OK? So, conveyor belts work because of friction if you were ice the object would just stay there, OK? That's it and I want to know how much work does this force do? In other word, what is the work... By the way this is static friction I mentioned static I said static and then I put that one there, what is the work done by this force? What is this work done by the static friction? The work done by kinetic friction is -Fd but that's not the case for static friction, to find the work done by static friction you're just going to use the regular work equations with no shortcuts because in this case it's actually going in the positive direction so you couldn't say negative because the angle is not negative the angle is 0 being not 180 it's 0 and I'm going to show you, the work done by a force is that force distance cosine of theta, friction's pulling me up and the displacement in that direction so the angle is 0, OK? But how much is friction? How much friction do I get? The key point in this problem to figure out how much friction you have is the fact that this moves at a constant speed 0 acceleration which means forces cancel, which forces here would cancel? Friction which is the force of the belt cancels out MGX, so I can say that friction static=MGX which in turn equals MG sine of theta, let's calculate real quick, MG sine 37 this is 30.... 30 Newtons, so this thirty goes here, what is the distance? You go 10 up the plane and we already mentioned the angle is 0, The angle is 0 that comes from here, so this work is 300 Joules and this should make sense as well, the static friction on the conveyor belt is (This is part C by the way how much work you do?) the static friction on the conveyor belt is pulling you up so there's 300 which is the same thing that gravity did, gravity did negative 300 to lift something you do a positive work and gravity does the negative of that, cool? That's it for this one let me know if you guys have any questions.

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A 50 g mass is lifted upwards 20 cm. How much work is done by the gravitational force?

A 360-kg piano slides 3.6 m down a 26 incline and is kept from accelerating by a man who is pushing back on it parallel to the incline.Determine the force exerted by the man.Determine the work done by the man on the piano.Determine the work done by the force of gravity.Determine the net work done on the piano. Ignore friction.

A grocery cart with mass of 14 kg is being pushed at constant speed up a flat 15 ramp by a force FP which acts at an angle of 17 below the horizontal.Find the work done by each of the forces ( mg, FN , FP ) on the cart if the ramp is 16 m long.

A physics professor is pushed up a ramp inclined upward at an angle 31.0 above the horizontal as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and chair is 86.0 kg . He is pushed a distance 2.55 m along the incline by a group of students who together exert a constant horizontal force of 601 N . The professors speed at the bottom of the ramp is 2.05 m/s .Use the work-energy theorem to find his speed at the top of the ramp.

You need to raise a heavy block by pulling it with a massless rope. You can either (a) pull the block straight up height h, or (b) pull it up a long, frictionless plane inclined at a 15
angle until its height has increased by h. Assume you will move the block at constant speed either way.Will you do more work in case a or case b? Or is the work the same in both cases?

A 20.0-kg crate sits at rest at the bottom of a 15.1 m -long ramp that is inclined at 36 above the horizontal. A constant horizontal force of 290 N is applied to the crate to push it up the ramp. While the crate is moving, the ramp exerts a constant frictional force on it that has magnitude 65.0 N.What is the total work done on the crate during its motion from the bottom to the top of the ramp?How much time does it take the crate to travel to the top of the ramp?

An
14.0-kg
package in a mail-sorting room slides
4.00 m
down a chute that is inclined at
56.0
below the horizontal. The coefficient of kinetic friction between the package and the chutes surface is
0.60.Calculate the work done on the package by friction.Calculate the work done on the package by gravity.Calculate the work done on the package by the normal force.What is the net work done on the package?

A gondola can carry 20 skiers, with a total mass of up to 1960 kg . The gondola ascends at a constant speed from the base of a mountain, at 2060 m , to the summit at 3555 m .How much work does the motor do in moving a full gondola up the mountain?How much work does gravity do on the gondola?If the motor is capable of generating 10 % more work than found in A, what is the acceleration of the gondola?

The cable of a crane is lifting a
750 kg
girder. The girder
increases its speed from 0.25 m/s
to 1.25 m/s
in a distance of
4.1
m.How much work is done by gravity?How much work is done by tension?

A 75-kg roofer climbs a vertical 7.0-m ladder to the flat roof of a house. He then walks 12 m on the roof, climbs down another vertical 7.0-m ladder, and finally walks on the ground back to his starting point.How much work is done on him by gravity as he climbs up?How much work is done on him by gravity as he climbs down?How much work is done on him by gravity as he walks on the roof?What is the total work done on him by gravity during this round trip?On the basis of your answer to part E, would you say that gravity is a conservative or nonconservative force? Explain.How much work is done on him by gravity as he walks on the ground?

What magnitude force is required to give a helicopter of mass M an acceleration of 0.10g upward?What work is done by this force as the helicopter moves a distance h upward?

In a certain library the first shelf is 15.5 cm off the ground, and the remaining 4 shelves are each spaced 33.5 cm above the previous one.If the average book has a mass of 1.40 kg with a height of 22.2 cm , and an average shelf holds 28 books (standing vertically), how much work is required to fill all the shelves, assuming the books are all laying flat on the floor to start?

The two ropes seen in the figure are used to lower a 255 kg piano 5.70 m from a second-story window to the ground.
You may want to review (Pages 214 - 218).How much work is done by each of the three forces?

A 66.5 kg hiker starts at an elevation of 1320 m and climbs to the top of a 2710 m peak.What is the hikers change in potential energy?What is the minimum work required of the hiker?Can the actual work done be greater than this?Explain.

A 76.0-kg firefighter climbs a flight of stairs 21.0 m high. How much work is required?

A 5 kg mass is pushed across a floor with a coefficient of kinetic friction of 0.35. If the force pushing the mass is 70 N, how much work is done after 5 s by:(a) the pushing force(b) friction after(c) the normal force(d) gravity

An object released from rest at the top of a 30.0° incline slides down the incline to the bottom of the incline. During this motion the work done on the block by the friction force on the block isA) positiveB) negativeC) zero

A box is pulled by a force F up a ramp that is inclined at 37° above the horizontal. The direction of this force is 60.0° above the horizontal. The force has magnitude 60.0 N. The box travels a distance of 5.00 m along the surface of the ramp. How much work does the force F do during this displacement of the box?A) 300 JB) 150 JC) 240 JD) 276 JE) None of the above answers

A block of mass 10.0 kg slides 16.0 m down a 36.9° incline, from point A at the top of the incline to point B at the bottom. As the block moves from point A to point B, the surface of the incline exerts a constant friction force that has magnitude 42.0 N.As the block moves from A to B, how much work is done on it by the friction force? (Be sure to indicate whether the work is positive or negative).

A block of mass 10.0 kg slides 16.0 m down a 36.9° incline, from point A at the top of the incline to point B at the bottom. As the block moves from point A to point B, the surface of the incline exerts a constant friction force that has magnitude 42.0 N.As the block moves from A to B, how much work is done on it by the gravity force? (Be sure to indicate whether the work is positive or negative).

A 1.75 m tall person lifts a 1.65 kg book off the ground so it is 2.15 m above the ground.(a) What is the potential energy of the book relative to the ground?(b) What is the potential energy of the book relative to the top of the persons head(c) How is the work done by the
person related to the answers in parts A and B?

A ramp is inclined at 36.9° above the horizontal. A block with mass 0.500 kg is pulled up the ramp by a force F. The block starts at point A at the bottom of the ramp and ends up at point B at the top of the ramp. The distance from A to B, measured along the ramp is 5.00 m. The work done on the block by gravity as the block moves from point A to point B is
A) +24.5 J
B) -24.5 J
C) +19.6 J
D) -19.6 J
E) +14.7 J
F) -14.7 J
G) None of the above answers

A block with mass 5.0 kg slides a distance of 6.0 m from point A at the top of a ramp to point B at the bottom of the ramp. The ramp is inclined at 53° above the horizontal. For the displacement of the block from A to B the work done on the block by gravity is
A) 177 J
B) -177 J
C) 235 J
D) -235 J
E) 294 J
F) -294 J
G) zero
H) None of the above answers

When an object is moved from rest at point A to rest at point B in a gravitational field, the net work done by the field depends on the mass of the object and
1. the nature of the external force moving the object from A to B.
2. the velocity of the object as it moves between A and B.
3. both the positions of A and B and the path taken between them.
4. the path taken between A and B only.
5. the positions of A and B only.

A boy in a wheelchair (total mass 47.0 kg) has speed 1.40 m/s at the crest of a slope 2.60 m high and 12.4 m long. At the bottom of the slope his speed is 6.20 m/s. Assume air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N. Find the work he did in pushing forward on his wheels during the downhill ride.

Spiderman, whose mass is 80.0 kg, is dangling on the free end of a 12.0-m-long rope, the other end of which is fixed to a tree limb above. By repeatedly bending at the waist, he is able to get the rope in motion, eventually getting it to swing enough that he can reach a ledge when the rope makes a 60.08 angle with the vertical. How much work was done by the gravitational force on Spiderman in this maneuver?

A 76.0-kg painter climbs a ladder that is 2.75 m long leaning against a vertical wall. The ladder makes an 31.0° angle with the wall. How much work does gravity do on the painter?

In 1990, Walter Arfeuille of Belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth. (a) How much work was done on the object by Arfeuille in this lift, assuming the object was lifted at constant speed? (b) What total force was exerted on Arfeuille’s teeth during the lift?

The record number of boat lifts, including the boat and its ten crew members, was achieved by Sami Heinonen and Juha Räsänen of Sweden in 2000. They lifted a total mass of 653.2 kg approximately 4 in. off the ground a total of 24 times. Estimate the total work done by the two men on the boat in this record lift, ignoring the negative work done by the men when they lowered the boat back to the ground.

A box with mass 5.00 kg is pulled up a 36.9° incline by a constant force F that has magnitude 75.0 N and that is parallel to the incline. The distance along the incline from the bottom to the top is 6.00 m. During the motion of the box, the surface of the incline exerts a constant friction force fk = 18.0 N on the box, in a direction opposite to the motion.For the motion from the bottom of the incline to the top, how much work is done by each of the following forces? In addition to giving the magnitude of the work, be sure to indicate whether the work done is positive or negative.(i) the force F that pulls the box(ii) the friction force fk(iii) the gravity force (iv) the normal force

A box of weight w = 2.0N accelerates down a rough plane that is inclined at an angle 30° above the horizontal. The normal force acting on the box has a magnitude n = 1.7 N, the coefficient of kinetic friction between the box and the plane is μk=0.30, and the displacement of the box is 1.8 m down the inclined plane.What is the work Ww done on the box by the weight of the box?

A box of weight w=2.0N accelerates down a rough plane that is inclined at an angle Φ =30° above the horizontal, as shown (Figure 6) . The normal force acting on the box has a magnitude n=1.7 N, the coefficient of kinetic friction between the box and the plane is μk=0.30, and the displacement d? of the box is 1.8 m down the inclined plane.Part a. What is the work Wn done on the box by the normal force?Express your answers in joules to two significant figures.Part b. What is the work Wfk done on the box by the force of kinetic friction?Express your answers in joules to two significant figures.

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