a-force-f-of-strength-20-n-acts-on-an-object-of-mass-3-kg-as-it-moves-a-distance-of-4-m-if-f-is-perp

Question

A force F of strength 20 N acts on an object of mass 3 kg as it moves a distance of 4 m. If F is perpendicular to the 4 m displacement, the work it does is equal to
(A) oJ
(B) 60 $\mathrm{J}$
(C) 80 $\mathrm{J}$
(D) 600 $\mathrm{J}$

Winnie Wu · Accepted Answer

So in order to think about this problem, we need to figure out the Adam and said we have been given. So we know that there&#x27;s a force off funny Newton and it is smoothing an object that has a mass and off three kilograms. And we know that so far the after husband displaced So it has a displacement stay off four meters and since pose force and displacement are vouchers and we need to figure out the direction off the vouchers and we know we say that displacement is going forward, then the forces perpendicular to the disc movement and divide the direction in Seda. We&#x27;ll see that the force has a direction off 90 degrees, since it&#x27;s perfect decorative displacement. And then finally we come to the work, just the unknown the same. No need to solving this problem and to help yourself the work. We need to find an equation that we can utilize the elements we have so far and plug it into the equation. So the best thing that applies here would be this one work because to Forrest Times displacement 10 times, coz I didn&#x27;t say that so and if we plan and say that First, which is 90 degrees. Who realized that co sign say that Coz I 90 degrees because 20 So in the end, we can see that when forces for been dicker to the div action that the after this morning the work that is done by the four slipping zero Joe&#x27;s and that is the ends of the problem.

Ze-Han Lee · Answer

right. So in this problem, we have object and the object has full meters displacement. So suppose this is it. This basement, which is four meters and the forces that perpendicular to this displacement half this is ax this out. So according to the definition of the work, work equal after an ex. So this is a vector product vector product. Okay, so this gives us zero. So the work down by this force is zero.

Ze-Han Lee · Answer

All right. So in this problem, we know the the false factor is full. I plus c j where i cz pointing to the ex direction the J&#x27;s pointing to the white direction. And we also know that this basement of actor the equal three i minors to J. Okay. And for part A, we want to find out, see if the final work is zero. So we just used the products f D. So in this case, it to 12 minus two seats, right? Zero we can find out. See equal six if a party if the work is ah, 17 Jews. So? So? So we have 12 minus two seed equal 17. So consult for C equal to negative 2.5 field too. Okay. And for Percy, if the work is negative 18 Jos 12 minus to see equal. Negative 18. So see, Finally equal 15 Newton.

Averell Hause · Answer

So we have a force vector and it&#x27;s this&#x27;s equaling 2.0. I had plus 4.0 j Hat Newtons and then we can find the A magnitude of this vector two squared plus four squared to the 1/2 power and this is giving us to radical. Five. Nunes. We can say that data is going to be equal to tangent. Negative one of four over too. So this will be our 10 of two. We know that the distance vector or the displacement vector would be equal to 1.0. I had plus 5.0 jae ha and the unit would be meters. Finding the magnitude of this would be one squared plus five squared. So this is giving us radical 26 meters. And so, in order to solve for party party, they want us to solve it the classical or the using the angle so f d co sign a fada and this would be a radical move rather. Rather, we could say to radical five Nunes times radical 26 meters and then co sign of attendant, Our 10 of five minus our 10 of two. And so work is going to be equal to 22 jewels now for part B, they&#x27;re simply asking us to find it the dot product way. So this will be path of X d of X plus half of why d of why. And this is giving us to Nunes Times one meter plus four Newtons times five meters and again, this is equal in the same answer of 22 jewels. Yeah, so you simply have to solve it both ways in order to prove that no matter how you solve it, you&#x27;re of course, getting the same answer. That is the end of the solution. Thank you for watching.

Hong Joo Ryoo · Answer

So to find the work done, you would have to integrate force with respect to our position. All since our components do not have anything to do with their position, Will Rosa&#x27;s multiply position The components do not have a position factor in them yet, and now they do. And that isn&#x27;t work. So to find the total work done, you will have to take the final value putting plug it in and then subtract our starting value or our initial value. Okay, great. Let&#x27;s find the orthe final value by plugging in five and 25. So plugging in five to our well X component. We have negative 20 and putting in 25. You will have 500. Now you&#x27;re subtract our initial value. It&#x27;s our initial value. Plugged in would be welp zero and 10 plugged in will be 200 distributing the well, its attraction sane. The zero will cancel out, leaving us with negative 24 r x component and 300 for away component. No, this is the work done in what the Y direction and the X direction. But the total work I would just be to add these components up, superposition said 300 minus 20 is going to be cool too. 280. And that is our total work.

Donald Albin · Answer

we have a force that varies with X. And I wrote it down up here. And ah, we&#x27;re given the value of A which is here noting noting that everything is in the S I based units. Um so ah, work is the integral of the ah force, which is a X to the 1/2 power um, D x. And that&#x27;s gonna go from 0 to 3. So we take the integral of this that&#x27;s going to give us X to the three halves power. It&#x27;s still over a and then we divide by three has, but that&#x27;s the same as multiplying by 2/3. And then X is going to go from zero 23 Now, I noticed that when x zero, I&#x27;m gonna be multiplying by zero. And so that&#x27;s gonna give me zero. So I only have to worry about the final state of three. So 2/3 times nine and 5/10 and then I have to worry about the three to the three halves power. And so again, I didn&#x27;t write minus 2/3 a time zero to the reactor, our power, because that&#x27;s gonna be zero. Um, so this is going to give us our answer for the work. I put that in my calculator, and I got a work of 30 three. Jules. Okay. Part B says now we got to go from 6 to 9 meters, so work is 2/3 times a X to the three halves. Power from six to nine meters. So that&#x27;s going to be 2/3 times 9.5. Um, Max is nine to the three. House power nine to the three halves. Power minus 2/3. Got a subtract the initial state of this one and the initial state with six to the three halves power. I&#x27;m going to put all of this in my calculator. I&#x27;m gonna pause my recording while I put it in, and then I will get right back to you. All right. I did that and I got an answer of 78 jewels. Now, I didn&#x27;t take this into the calculator directly. I actually factored out the 2/3 times 9.5. Um and then I put it into the calculator, but you could put it in the calculator directly as well. And so, um, these our our our answers Thio A and C o man, I didn&#x27;t realize that there was another one. So a see. All right. Looks like I forgot to do be So let me go to go back and do be And I definitely ran out of room on this. So I&#x27;m gonna erase, supposed to do from 3 to 6. So that&#x27;s whoops. Did not want to do that. So I&#x27;m gonna do this quickly. It is. And I&#x27;m gonna factor it out so that you can see it. So in part B, it&#x27;s too. Oh, my goodness. Okay, 2/3 times 9.5, um, times. Ah, this is going from 3 to 6. So that would be six to the three halves. Power minus three to the three halves. Power. Okay, that&#x27;s going to give us our answer for part B. Um, so I&#x27;m gonna pause again and, ah, come back to you. Okay, So I put that into my calculator, and I got 60. Jules. All right, So our three answers are 33 jewels, 60 jewels and 78 Jules

Problem

Under the influence of a force, an object of mass…

Problem 1 Easy Difficulty

A force F of strength 20 N acts on an object of mass 3 kg as it moves a distance of 4 m. If F is perpendicular to the 4 m displacement, the work it does is equal to
(A) oJ
(B) 60 $\mathrm{J}$
(C) 80 $\mathrm{J}$
(D) 600 $\mathrm{J}$

Answer

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