Drake Glau Posted December 19, 2010 Report Share Posted December 19, 2010 (edited) There's a biology and chemistry help thread, figured a physics one would be appreciated by a few people Hoping some people will check here and help people who have questions, I'll try my best but my knowledge of physics does not go very far honestly and it's not my best science Edited December 19, 2010 by Drake Reply Link to post Share on other sites More sharing options...
dessskris Posted December 28, 2010 Report Share Posted December 28, 2010 I guess I would start off What is the relationship between Force and Velocity? I've got some questions about it but I guess I shall ask one only and do the rest myself.An object of mass m is pulled along a horizontal track by a constant horizontal force Fp. The frictional force between the track and the object is Ff. After the object has been moved a distance d from rest, its speed will beA. (Fp - Ff)d/2mB. 2d(Fp - Ff)/mC. sqrt[2(Fp - Ff)d/m]D. sqrt[(Fp - Ff)d/2m]and explain why? I guess it has got something to do with Ek = mv2/2 but I don't know. Reply Link to post Share on other sites More sharing options...
Drake Glau Posted December 28, 2010 Author Report Share Posted December 28, 2010 (edited) v = atd=vt+1/2at², the v is initial velocity which is 0 so...d = vt = 1/2 at²d = 1/2 a(d/v)² = 1/2 a d²/v²a = 2v²/dF=maF=m(2v²/d)F/m=2v²/dFd/m=2v²Dividing by two, multiply by 0.5, same thing so(1/2)Fd/m is the same as Fd/2m soFd/2m=v² thereforev=sqrt[Fd/2m]Friction and the force you're pushing with work in opposite directions ( ) so your F=Fp-FfAnser should be D. v=sqrt[(Fp-Ff)d/2m] Edited December 28, 2010 by Drake 2 Reply Link to post Share on other sites More sharing options...
dessskris Posted December 28, 2010 Report Share Posted December 28, 2010 Thanks, I get that now!Two 10 kg blocks on a smooth horizontal surface are tied together. They are accelerated by a horizontal force of 30 N which acts as shown below:╔╗ -----connecting rope----- ╔╗ ----> 30 NIf frictional effects are negligible, what is the tension in the connecting rope?A. 30 NB. 15 NC. 10 ND. 0 NI think the answer should be B. is that correct?A massive block is supported by a thread A from the ceiling, and another thread B is attached at the bottom, like the picture belowwithout the handA downward force is applied to string B. This force can be a steady pull or a suden jerk.Which one of the following statements about the force is true?A. A sudden jerk breaks A, a slow pull breaks B.B. A sudden jerk breaks B, a slow pull breaks A.C. A steady pull or a sudden jerk breaks A.D. A steady pull or a sudden jerk breaks B.I think it should be B or D Which and why? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted December 28, 2010 Author Report Share Posted December 28, 2010 Thanks, I get that now!Two 10 kg blocks on a smooth horizontal surface are tied together. They are accelerated by a horizontal force of 30 N which acts as shown below:╔╗ -----connecting rope----- ╔╗ ----> 30 NIf frictional effects are negligible, what is the tension in the connecting rope?A. 30 NB. 15 NC. 10 ND. 0 NI think the answer should be B. is that correct?I feel like it's just A honestly. Since there's no friction, you pull on the one mass with the 30N so it's then pulling on the rope with 30N also because there is no friction. This could be wrong though, you're asking questions that was part of the first year of physics that I skipped A massive block is supported by a thread A from the ceiling, and another thread B is attached at the bottom, like the picture belowwithout the handA downward force is applied to string B. This force can be a steady pull or a suden jerk.Which one of the following statements about the force is true?A. A sudden jerk breaks A, a slow pull breaks B.B. A sudden jerk breaks B, a slow pull breaks A.C. A steady pull or a sudden jerk breaks A.D. A steady pull or a sudden jerk breaks B.I think it should be B or D Which and why?This question is ridiculously vague? Or I'm super complicating it o.O I'd pick B personally, I don't know why I would pick B...but I would Reply Link to post Share on other sites More sharing options...
Tensai Posted December 29, 2010 Report Share Posted December 29, 2010 (edited) Thanks, I get that now!Two 10 kg blocks on a smooth horizontal surface are tied together. They are accelerated by a horizontal force of 30 N which acts as shown below:╔╗ -----connecting rope----- ╔╗ ----> 30 NIf frictional effects are negligible, what is the tension in the connecting rope?A. 30 NB. 15 NC. 10 ND. 0 NI think the answer should be B. is that correct?B is correct, I believe.Force = Mass * Accelerationand because the two blocks are connected, they will have the same acceleration. So if there's 30 N acting on the system with a total mass of 20 kg, a = 1.5 m/s^2Then, the tension in the connecting rope is the force being exerted to move only the second box with the same acceleration. So: 10 kg * 1.5 m/s^2 = 15 NNo idea about the second one, sorry. Common sense seems to tell me it's C, but maybe that's just me and besides I learned long ago not to trust common sense when it comes to physics Edited December 29, 2010 by Tensai 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted December 29, 2010 Report Share Posted December 29, 2010 Thanks people!For that steady pull/sudden jerk, right, I think since the block is massive, it must be very very heavy, so the Weight should be great. Thus the Tension in thread A should also be very great.Now the Force we apply to thread B could be a sudden jerk or a steady pull. IMO, according to my common sense: if the force is a sudden jerk, this force should not be great. Thus F<T and it will not break thread A but it will break thread B. If the force is a steady pull, this force may gradually increase. Thus F may approach T, it may break thread A and idk what will happen to thread B.But I think it still depends on the amount of force applied to thread B.On the other hand, I somehow think both will just break thread B (answer D) because the block is massive.Shall we conduct an experiment to answer this question...? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted December 29, 2010 Author Report Share Posted December 29, 2010 Thanks people!For that steady pull/sudden jerk, right, I think since the block is massive, it must be very very heavy, so the Weight should be great. Thus the Tension in thread A should also be very great.Now the Force we apply to thread B could be a sudden jerk or a steady pull. IMO, according to my common sense: if the force is a sudden jerk, this force should not be great. Thus F<T and it will not break thread A but it will break thread B. If the force is a steady pull, this force may gradually increase. Thus F may approach T, it may break thread A and idk what will happen to thread B.But I think it still depends on the amount of force applied to thread B.On the other hand, I somehow think both will just break thread B (answer D) because the block is massive.Shall we conduct an experiment to answer this question...?Pfft, experiments...I think since the block is massive it acts as a mass between whatever is pulling down and the block. So a quick jerk would apply a ton of force between the massive block and your hand (just cause...) which would break the bottom thread (B?). A gradual tug would increase the force applied to the rope, increasing it's tension, but the force would then move through the block so to speak and to the next thread (A?). From doing this it now has the force of the block AND your hand whereas the bottom thread only has your hand's force. Assuming both threads are the same then the gradual pull would cause the top thread to snap first as the force slowly increases from your hand but the extra block mass applies just a little more to push A over the edge before B. You like math soooooLet's say I'm trying to get to a number, 5. Gradually pulling I just add 1 each second, both thread A and B start at 0, however thread A has an extra 1 at the start so it starts at 1 A in 4 seconds has reached 5, B has reached 4, hope this made sense, it made sense in my head. The sudden jerk would just immediately add 5 to B, snap it, causing the force to stop and never get to A. 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted December 29, 2010 Report Share Posted December 29, 2010 Two forces of magnitudes 7 N and 5 N act at a point. Which one of the following is not a possible value for the magnitude of the resultant force?A. 1 NB. 3 NC. 5 ND. 7 NI don't have any idea o__oI think possible ones would obviously be 2 N and 13 N. Then......?It really depends on the direction right? I personally think it's A because I think the possible resultant force would be between 2 N and 13 N inclusive, and 1 N is not in the range.Anybody has a better answer and explanation? Reply Link to post Share on other sites More sharing options...
Tensai Posted December 29, 2010 Report Share Posted December 29, 2010 (edited) Two forces of magnitudes 7 N and 5 N act at a point. Which one of the following is not a possible value for the magnitude of the resultant force?A. 1 NB. 3 NC. 5 ND. 7 NI don't have any idea o__oI think possible ones would obviously be 2 N and 13 N. Then......?It really depends on the direction right? I personally think it's A because I think the possible resultant force would be between 2 N and 13 N inclusive, and 1 N is not in the range.Anybody has a better answer and explanation?That seems right. If the forces are acting in the exact opposite direction the net force would be 2N, as you noted, and in the same direction 13N. Then, with any angle between the two that's between 0 and 180 degrees, if the two forces and the resultant force are arranged to form a triangle (as they would be to solve for the resultant force), one property of triangles is that any two sides add up to more than the 3rd side. So then 5N + (resultant force) > 7N and thus the resultant force can't be 1N.So yeah. You were right Edited December 29, 2010 by Tensai 2 Reply Link to post Share on other sites More sharing options...
dessskris Posted December 30, 2010 Report Share Posted December 30, 2010 Thanks, that makes sense.When an object of mass m is suspended from the end of a vertical spring, it extends a distance e. The object is now pulled down a further small distance a, and then released. The maximum kinetic energy of the mass isA. mg a2 / 2eB. e a2 / 2mgC. mg e2 / aD. mg e2 / 2aEk is ½mv2, right? Then? How is that related to e, a and g?I think B is definitely wrong because the m should be a numerator. I think C may be wrong, too. So it's either A or D but which one?Also, does anybody know any equation which shows the relationship between power, mass and speed? Reply Link to post Share on other sites More sharing options...
Eydie Posted December 30, 2010 Report Share Posted December 30, 2010 (edited) A massive block is supported by a thread A from the ceiling, and another thread B is attached at the bottom, like the picture belowwithout the handA downward force is applied to string B. This force can be a steady pull or a suden jerk.Which one of the following statements about the force is true?A. A sudden jerk breaks A, a slow pull breaks B.B. A sudden jerk breaks B, a slow pull breaks A.C. A steady pull or a sudden jerk breaks A.D. A steady pull or a sudden jerk breaks B.I think it should be B or D Which and why?I think the correct answer, as said, is B. We actually did the experiment in Year 10 - and our teacher explained as the follows:When there is a sudden jerk, the tension in B is so great that it breaks. String A does not break in this instance because the mass in between A and B has inertia, or you can think the definition of mass in terms of an ability to resist change.When there is a steady pull, tension in A is greater than that in B, because A has to sustain the pull as well as the weight of the mass, whereas B only need to sustain the pull.Sorry for the repetition, but I think this is simpler! Edited December 30, 2010 by Eydie 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted December 30, 2010 Report Share Posted December 30, 2010 I think the correct answer, as said, is B. We actually did the experiment in Year 10 - and our teacher explained as the follows:When there is a sudden jerk, the tension in B is so great that it breaks. String A does not break in this instance because the mass in between A and B has inertia, or you can think the definition of mass in terms of an ability to resist change.When there is a steady pull, tension in A is greater than that in B, because A has to sustain the pull as well as the weight of the mass, whereas B only need to sustain the pull.Sorry for the repetition, but I think this is simpler!Thanks! Just this morning I saw another question like this:A mass is supported from the ceiling by a string. A force is applied to a second string connected to the lower side of the mass. When the force is strong and sudden, the string breaks at B. When the force is initially small and gradually increases over a period of time, the string finally breaks at position A.[similar picture]Which one of the following Laws explains why the string breaks in one position or the other?A. Newton's First LawB. Newton's Second LawC. Newton's Third LawD. The Conservation of MomentumAnd I was like...oh **** the answer of the previous question is here..hahaDo you think the answer of the question is C (Newton's Third Law)? I am not good at this kind of thing o_o Reply Link to post Share on other sites More sharing options...
genepeer Posted December 30, 2010 Report Share Posted December 30, 2010 Thanks, that makes sense.When an object of mass m is suspended from the end of a vertical spring, it extends a distance e. The object is now pulled down a further small distance a, and then released. The maximum kinetic energy of the mass isA. mg a2 / 2eB. e a2 / 2mgC. mg e2 / aD. mg e2 / 2aEk is ½mv2, right? Then? How is that related to e, a and g?I think B is definitely wrong because the m should be a numerator. I think C may be wrong, too. So it's either A or D but which one?Also, does anybody know any equation which shows the relationship between power, mass and speed?Still figuring out the first question. And for the second one, do you mean mass or force, speed or velocity? I know one for Power, Force, Velocity.Power = Work Done / Time = Force * Displacement / Time = Force * Velocity. Reply Link to post Share on other sites More sharing options...
dessskris Posted December 30, 2010 Report Share Posted December 30, 2010 do you mean mass or force, speed or velocity? I know one for Power, Force, Velocity.Power = Work Done / Time = Force * Displacement / Time = Force * Velocity.Nah the question is asking:Sand is poured vertically on to a moving conveyor belt at a steady rate so that the mass of sand landing per unit time is m. The conveyor belt maintains a constant speed of v. The sand slips on the conveyor belt until it also reaches the speed v.[figure which I am too lazy to scan]The minimum power required to keep the conveyor belt moving at constant speed while the sand is falling on it will beA. zero.B. m v.C. ½ m v².D. m v².I think I have been asking too many questions haha I still have 100 more to do (they are all MCQ) Reply Link to post Share on other sites More sharing options...
genepeer Posted December 30, 2010 Report Share Posted December 30, 2010 (edited) Ok I'll make an educated guess and pick A. Since, mg = ke (force proportional to extension), C & D will have Ek = ke3/a or ke3/2a. That cube doesn't feel right. B is wrong as expected. Plus energy's units can be expressed as Nm, and B's unit don't simplify to that. Usually, you'd solve this question by equating maximum kinetic energy to the elastic (potential) energy gained by extending the spring. I see no reason why A is the answer but it feels less wrong than the others. Edited December 30, 2010 by Gene-Peer Reply Link to post Share on other sites More sharing options...
Eydie Posted December 30, 2010 Report Share Posted December 30, 2010 Thanks, that makes sense.When an object of mass m is suspended from the end of a vertical spring, it extends a distance e. The object is now pulled down a further small distance a, and then released. The maximum kinetic energy of the mass isA. mg a2 / 2eB. e a2 / 2mgC. mg e2 / aD. mg e2 / 2aEk is ½mv2, right? Then? How is that related to e, a and g?I think B is definitely wrong because the m should be a numerator. I think C may be wrong, too. So it's either A or D but which one?I think it's A.Recall/picture a Hooke's law experiment graph: Force F on y-axis, extension delta x on x-axis, a straight line going through the origin with slope k (k = the spring constant), and the area under the graph is the work done. Work done to extend the spring = the kinetic energy possessed by it once released.Work = (ka.a)/2 (working out the area of the triangle/area under the graph, where the x-coord is a, y-coord is ka (as F = k delta x and delta x here is a)= (ka^2)/2Recall again that as F = k delta x.When x = e, the force applied to extend that is the weight of the mass, or weight =mgThus it can be written as F = k e OR mg = k eRearrange gives k = mg/eNow substitute k = mg/e into (ka^2)/2yields (mga^2) / 2ewhich is option A(Someone please confirm : ) ) 2 Reply Link to post Share on other sites More sharing options...
genepeer Posted December 30, 2010 Report Share Posted December 30, 2010 (edited) do you mean mass or force, speed or velocity? I know one for Power, Force, Velocity.Power = Work Done / Time = Force * Displacement / Time = Force * Velocity.Nah the question is asking:Sand is poured vertically on to a moving conveyor belt at a steady rate so that the mass of sand landing per unit time is m. The conveyor belt maintains a constant speed of v. The sand slips on the conveyor belt until it also reaches the speed v.[figure which I am too lazy to scan]The minimum power required to keep the conveyor belt moving at constant speed while the sand is falling on it will beA. zero.B. m v.C. ½ m v².D. m v².I think I have been asking too many questions haha I still have 100 more to do (they are all MCQ)Oh, that. It has something to do with changing momentum and that particular question was used as an example in our textbook. I never really understood it properly 'cause the book was horrible. If I remember correctly, then in one second, m mass of sand lands on the belt. This mass experience a (horizontal) velocity change of v. The change in momentum is mv and therefore the force exerted by the belt on the sand is mv (Newton's 2nd Law of Motion I think). In that second, the sand travels a distance of v. Therefore the work done in one second (Power) is mv2. So I think it's D. Edited December 30, 2010 by Gene-Peer 2 Reply Link to post Share on other sites More sharing options...
Eydie Posted December 30, 2010 Report Share Posted December 30, 2010 A mass is supported from the ceiling by a string. A force is applied to a second string connected to the lower side of the mass. When the force is strong and sudden, the string breaks at B. When the force is initially small and gradually increases over a period of time, the string finally breaks at position A.[similar picture]Which one of the following Laws explains why the string breaks in one position or the other?A. Newton's First LawB. Newton's Second LawC. Newton's Third LawD. The Conservation of MomentumAnd I was like...oh **** the answer of the previous question is here..hahaDo you think the answer of the question is C (Newton's Third Law)? I am not good at this kind of thing o_oI think it might be B. Because when explaining it, a = F/m (which is the Second Law) is used: when a force F is exerted on a mass, if the mass m is big, the acceleration a caused is small... (although I remember doing this question when we were learning the First Law) 1 Reply Link to post Share on other sites More sharing options...
dessskris Posted January 4, 2011 Report Share Posted January 4, 2011 Thank you very much people!More questions? [HL, Gravitation and Orbital Motion] A satellite is orbiting the earth in a circular orbit. Which one of the following properties of the satellite does not remain constant?A. Kinetic energyB. Gravitational potential energyC. Angular momentumD. VelocityI think A and B should be both constant since the R is not changing. So it's either C or D. What is angular momentum, though? How is it different from momentum? I think velocity should be constant but may be not due to acceleration and air resistance... (even though I think air resistance is negligible) Reply Link to post Share on other sites More sharing options...
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