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Snownfiqure below,Mass m = 51,0 kg (initia at rest) pushed distance 82.0 m OCross rouqn warehouse fioor applied force of FA 216 directed at an angl honzontal. Thc c...

Question

Snownfiqure below,Mass m = 51,0 kg (initia at rest) pushed distance 82.0 m OCross rouqn warehouse fioor applied force of FA 216 directed at an angl honzontal. Thc cocfficicnt of kinctic friction betwecn the @loor and the box 100_ Determinc the following- (For parts (a) through (d) give Your answer Lhc ncurcst mulupic 10.)30.09 Delow theroch surfecsorC done by the applied forceWon done by the forcegravilywotk done by the normal forceMonC done by the forcefrictionCakculatc thc nct work on thc box

snown fiqure below, Mass m = 51,0 kg (initia at rest) pushed distance 82.0 m OCross rouqn warehouse fioor applied force of FA 216 directed at an angl honzontal. Thc cocfficicnt of kinctic friction betwecn the @loor and the box 100_ Determinc the following- (For parts (a) through (d) give Your answer Lhc ncurcst mulupic 10.) 30.09 Delow the roch surfecs orC done by the applied force Won done by the force gravily wotk done by the normal force MonC done by the force friction Cakculatc thc nct work on thc box by finding thc suM all thc works donc by indiviquz iorcc WNet Now find the net work by first finding the net force on the box; then finding the work done DY this net force. WNet



Answers

A55-kg box is being pushed a distance of $7.0 \mathrm{m}$ across the floor by a force $\overrightarrow{\mathbf{P}}$ whose magnitude is $160 \mathrm{N}$. The force $\overrightarrow{\mathbf{P}}$ is parallel to the displacement of the box. The coefficient of kinetic friction is $0.25 .$ Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force.

To solve this question first, we need to draw the free bollock dagger. So let's show the free body diagram for the box, The four speed, which is a pulling force. And it is tempting to us. The I was in the direction they stated, the direction along the force acting is X. And the particular action is like in the way dissecting along the downward direction that is MG. And a normal reaction force accepting longer vertical direction and affection force X. Opposite to the direction of motion. And the friction force is if you're into the kinetic friction into uh normal reaction force. Okay. And let's say that displacement after boss is along the exchange with the devastation. Mm. So from Newton's law, the box is not moving along the vertical direction. So we can say the net force along the right direction as if you were in brazil. And from this we can say the man killed at the normal reaction force is equivalent to the weight, accept the situation and the mention of the friction forces guarantee milk a times often. Or you can say the magnitude of the friction forces recover into it. Look at times and no the warden by the normal reaction force is the Cuban people. The marriage that the normal reaction force that is empty into the displacement. That is s into the angle between the normal reaction force and the displacement. That is cause of 90° And cause of 90° 0. Therefore, the work done by the normal reaction force is a Cuban and 202 weeks. Now the work done by the gravity uh is secure and two MG into the discussion vector S. And two. Again. The angle between both of them is 90 degrees. Therefore, course of 90 degree. Therefore the work done by the gravity is also zero shoots. Now the world done by the 4th p. is secure rental. The manage of the first P. Into the displacement. As Into the angle between them. That is cause of 0°.. Therefore the work done by the pulling forces clean to S. And if we substitute the value the work done by their forces. If you're into The magnitude of the force that is 150 newtons Into the displacement, that is seven meters. So the work done by the force of the community 1050 Jews. Or if you write this in two significant figures, the work done by the forces that you were into 1.1 into 10 days, three troops. Now the work done by the friction is a covenant. The magnitude the friction force that is my okay into MG. And to the displacement. That is S into because of the angle between the four center displacement is one. Hdgs Are the work done by the friction forces preparing to minus of zero point 25. That is the kinetic friction into the mass that is 55 kg In 29.8 m for 2nd square And 27 m. Or the work done by the friction forces -943 125 jewels are in two significant figures. The work done by the friction forces -940 juice.

Alrighty. So we have this five kilogram box that's sitting on a horizontal surface. Here it is. M equals 5.0 kilograms. And there's a coefficient of kinetic friction on this surface, and that's equal to 0.50 And the horrors that we there's a horizontal force that pulls the box here and that horizontal force. It's not given to us. We want to. And keep in mind, though, that this is, um, not given to us, and it's gonna keep it the equal sign that we have to know. One thing at the velocity when this thing is in motion is equal to a constant. So the net Force is gonna be equal to zero. We need to find the work done by a few things. Euro Last thing. We should definitely also mention the same gonna move a distance here. Distance D We may make some more space here. My coefficient of static friction. I can move this over here and take my displacement vector back here. Cool. Oh, there we go. And back to my marker Heikal so d equals they give it to us and centimeters. So let's keep it like that for now 10 centimeters. Let's figure out the work done by the supplied horizontal force by the Frictional Force and by the Net Force. So three different things we got to figure out here. So let's try to figure out the one that work done by Ah, my friction. Right. Do that first. So we know that the work done by friction is going to equal. It's always negative here because that frictional force is gonna be opposite of our motion. I won't put a negative in. Yeah, logistician you guys what I mean here. So number frictional force is equal to the normal force multiplied by the coefficient of friction. Normal force here is gonna be equal to N G and the coefficient of friction music camp and removing a distance d just 10 centimeters. Now remember, this is the force multiplied by the distance and then multiplied by the angle between the vectors. That's your normal work equation that we use. This guy right here is negative one. And as results, you papa negative one of everything here. So have a negative M g m u sub k times D, which is equal to a negative five times 9.8. Not gonna bother my units here. That's the gravity I pointed. One multiplied by the coefficient of friction. And for the 10 centimeters, seven multiplied by 70.1. And then we will get the work done by friction. And that is equal to negative. 2.45 Excuse me? My marker had his little freak out there. Jules. All right, that's the work done by friction. All right? No. Remember, this thing is moving at a constant velocity. All right? So that means the work total has to be equal to zero, and work total is going to be equal to your work off your friction multiplier. Um, and then plus the work done by the, um, Applied Force. And it's equal to zero because work totals equal to zero. What does this mean? This means that these guys are equal to an equal in magnitude opposite in sign. So you work done by your applied force is going to now be a positive 2.45 Jules

For this problem, we will be looking at the work energy theorem and how it applies to several different forces in the system. The first force we will be looking at is this pushing force that is applying 130 Newton's on the box, which is causing it to slide along the floor. The work done by this force is equal to the magnitude of the force times the distance that the box travels, which is 130 Newton's times the five metres, which gives 650 jewels. Notice that the work done by this force is positive because the force and the displacement of the box are pointing in the same direction. The second force will be looking at is the fresh and force. This the work done by this force is going to be the negative of the friction force times the distance at the box travels. In this case, because the friction forces opposing the motion of the box, the work done by this force is negative. The first enforce can be found by using by multiplying the coefficient refreshing times normal force and the normal forces simply just the weight of the box. It's mass times gravity, which is 40 kg times 9 21 meters per second squared, which gives a normal force of 392 0.4 newton's to find the first enforce, we multiply the 392 0.4 Nunes by the coefficient of friction, which is 0.3 which gave 117 Yeah 0.72 newtons. Now we can plug back into the work formula which gives us that the work done by the friction forces negative 117.72 newtons times of five m of the box travels, which gives us negative 588 0.6 jewels. Now we can look at the work done by the normal force. This is simple because the law enforces pointing straight up while the box is only being displaced in the horizontal direction, which means that the work done by this force zero. Similarly, the work done by gravity is also zero because the force of gravity is pointing straight down and the blocks only moves left and right. Now we can think about the kinetic change in kinetic energy that the force experience during this movement. The change in kinetic energy, it's equal to the work non conservative done by all the forces in the system. In this case, both the pushing force of 130 Newton's and the frictional force are non conservative, which means that the change in kinetic energy is equal to the work done by the pushing force minus the work plus the work done by the frictional force, which gives us that. The change in kinetic energy is 650 jewels minus the 588 0.6 jules, which means that the change in kinetic injury is 61 point or jewels notice that it I subtracted 588.6 jewels because we found that that work done by the frictional force was negative since it opposed the motion. Finally, we can find the final velocity of the box. Since we know the change in kinetic energy is equal to 61.4. This must be equal to one half the math. The box times the final velocity squared minus one half the the mass of the blocks, times the initial velocity squared. We know that the initial velocity of the boxes zero since it started at rest, which can souls this term to zero. So we're left with 61.4 is equal to one half the mass of the box, which is 40 kg times the final velocity squared. This allows us to find the final velocity, which is 1.75 meters per second.

Hello, guys. Here we have a situation where the but it starts from rest and then moves due to a net force. The total displacement is the first. We're going to calculate the velocity When we have half off the total work, we can do it using kinetic energy. The work off the net force is equal to the variation off the kinetic energy. In the beginning, it zero So when the work reached half off the total dragnet IQ energy is also half off the total. So let's compare the final kinetic energy is and the half kinetic energy is thus the relation between these velocities is which is greater than half the final speed. Now we calculate the work done when the bots rich half its final speed. The total work is the work done when it rates half the final speed is comparing. Then we have. So the work done at this point is a quarter off the total work, which means it is less than half off the total work. Bye


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