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A few exam questions

  1. Jun 22, 2006 #1
    I just got back from doing my physics exam, didn't do so well, but I did well during the semester and I think I did well on a final summative project.
    This is gr. 11 by the way, so nothing to fancy.

    Anyway, I wrote down the questions i wasn't sure of, and I just want someone to help me out with the answers please:


    4 students timed(something) and got the times 1.85, 1.8, 1.9, and 2, what was the average?

    I know this looks really basic and all, the answer I got was 2, simply because of following the significant figure rules:

    1.85 + 1.8 + 1.9 + 2 = 7.55, accurate to no dec. place because of the number '2'.

    Therefore it's 8, divide by 4 = 2, accurate to one sig fig, wierd result yes, but that's how I did it, look alright?


    Which of the following demonstrates newton's 3rd law?

    turning a corner
    2 skaters pushing against one another and being sent backwards
    earth held in orbit by the sun
    friction of driving on a road(I think)
    car tires slipping

    I think it's the one with skaters, although in some sense the 3rd law can be seen anywhere, I also thought the earth one was possible, because the sun pulls the earth, and so the earth pulls the sun, I don't know, someone please tell me if I'm right, it wasn't the best question I know but...


    What is true about the gravitational field strength of earth?

    9.8 everywhere
    greatest at equator
    smallest at the top of mt. everest
    largest at the deepest spot in the ocean
    largest at the poles

    It's one of the last two I'm pretty sure, though really only the last makes sense, sort of since the earth bulges at the equator, and the radius is smaller at the poles. Though none are really true, because you can go higher than mount everest, and lower than the ocean floor and the poles, but what would you say?


    Which of the following does not contribute to kinetic friction?

    coefficient of friction
    normal force
    applied force
    gravitaitional field strength

    I guessing normal force, though I'm not sure; I was absent when we did most friction related stuff, could someone correct and explain why?


    What's the angle of the reflected ray with an incident ray of 34 degrees?


    I'm thinking 56, because we were told that reflected rays make a 90 degree angle with the incident ray... so 90 - 56 = 34, am I right?


    What's the critical angle of ethyl alchohol(index of refraction 1.36) in air(index of refraction 1.0003)? Or was it the other way around?


    I'm thinking 47.3

    sin(critical angle) = n_2/n_1
    theta = 1.0003/1.36 sin(inverse)
    = 47.35 or something like that, am I right?

    That's all for now, thanks, someone please help soon.
  2. jcsd
  3. Jun 22, 2006 #2


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    I'm afraid not, you should only round at the end of the calculation; not mid calculation. Therefore, you should obtain;

    [tex]\frac{7.55}{4} = 1.8875 = 1.89 \; (\text{3sf})[/tex]
    All the red ones are examples of newton's third law. You can think of friction being the force the tyres exert on the earth, if the earth didn't push back the wheels would just spin and you wouldn't go anywhere. EDIT: Actually, they're all examples of newton's thrid law as you say, as newton's third law is valid everywhere. But the example were it would be most obvious is the skaters as you correctly said.

    I would agree with you there. Besides, once you go below the surface of the earth (including any water), Newton's law of gravitation breaks down.
    I'm afraid your guess is wrong by a country mile. What is the equation forthe kinetic frictional force?
    I'm afraid this is also wrong. What is the law of reflection? Probably the first rule you learned when considering mirrors.
    I would agree with you there. Although none of the given answers are technically correct because 47.35 = 47.4(3sf)
    Last edited: Jun 22, 2006
  4. Jun 22, 2006 #3
    (this is in response to hootenanny, jsut in case someone else posts before I'm done)

    Err, yes I considered that for the first one as well, the teacher showed us a similar example where I think the way was like what I did, so really the answer is wrong, I figured, but someone else(with a bit higher average than me) got the same answer, I'll see how my teacher marks it, I agree with you, I was originally going to do that, and of course my first impreesion was right(changing my first impression cost me several more marks on the test too...)

    The second one, yes, I think it sounds right, but turning a corner applies this too right? (applying a force outwards I believe, and the road responds by pushing you inwards, though I think there was another bit in that choice... car tires slipping(on an icy surface I'm pretty sure was the other part) maybe as well, well the 3rd law is present anywhere just about. Hmm, if you are applying a force to the ice(backwards) to move, then shouldn't the ice still push you forwards? I think I'm missing something here, could you clarify?

    The third one... I don't know, what did you mean by the gravitaion breaking down in the water?

    Fourth, ugh like I said, I was absent( I know I should've gotten the notes), I don't know about the equation of which you speak, and I no longer have my textbook to refer too, could you help me out here?

    Fifth, I heard the incident ray made a right angle with the reflected ray, what went wrong here?

    Last one, yeah, I didn't think anything was right either, that one was just the closest so... yeah.

    Oh, here's one more, I'm pretty sure about it, just need to make sure:

    A rocket reaches an altitude of 4.50 X 10 ^ 2 m in 4.0s, what is the accleration, what is the velocity?

    Here's what my friend showed me(for the first part):

    d = v_1(t) + 0.5(a)(t ^ 2)

    I was going to use that one too...

    v_1 = 0 therefore

    d = 0.5a(t^2)
    2d =a(t^2)
    a = (2d)/(t^2)
    = ((2)(450))/4.0^2
    = 900/16
    = 65.25 m/s^2

    Is that right?

    The velocity is easy:

    v = d/t
    = 450/4
    = 112.5


    Oh wait, is it the other one?

    v_2 = v_1 + t(a)?
  5. Jun 22, 2006 #4


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    I edited my post after some consideration (obviously while you were replying)
    The easiest way to think about this is when you are outside or on the surface of the earth, all the gravitational force created by all the molecules which make up the earth are all pulling in the same direction, towards the centre of gravity. Now, if you go below the surface, such as under water, the water molecules will be pulling you in all different directions, the ones above you will be pulling you up, the ones to either side will be pulling you sideways etc. Do you follow?
    I'm not happy with this question, becuase all the factors listed can affect the size of the frictional force. Were there no other constraints? For your reference the kinetic frictional force is given by [itex]F = \mu R[/itex], where R is the normal reaction force. This question is poorly worded and impossible to answer. I am dissapointed with your physics teacher if the question was presented this way.
    Thats wrong, thats what went wrong. Quite simply, the angle of reflection is equal to the angle of incedence. Where did you here this rule of yours?
    That's probably a type but the correct answer is 56.25 m.s-2
    I'm afraid thats wrong, the above formula assumes constant velocity, which we haven't got in this case. Your second formula would be the one to use;
  6. Jun 22, 2006 #5
    Woops yeah, the answer was 56.25... Also, what about the car on ice question, I'm a little curious, if the car applies a force backwards, the ice pushes forwards, but as we all know, driving on ice isn't easy, could you explain what else is going on here? I know you editted your post, but I'm just curious...

    And yes, I always thought the angle of relfection was equal to the angle of incidence, I'm not sure where the right angle notion came from exactly...

    Also, why do the particles in water pull you in different directions? The center of gravity is still beneath you, could you explain please?

    Also, I'm not entirely sure about the friction factors one.So the kinetic friction is a result of whatever greek character you used there, and the normal force? I thought mass played a role as well? There was also a free-body diagram with it, looking something like this:

    (consider O to be the object, and ignore dots)

    ..(friction) ------- O ------- (applied)

    Don't know if this helps any...

    And inside the object was velocity(in direction of applied) and I think the letter you used in that equation(for kinetic friction) was also inside. Also for that question, it might've said that greek letter instead of coefficient of friction for that option, not entirely sure if that helps at all either.
    Last edited: Jun 22, 2006
  7. Jun 22, 2006 #6
    Not sure if time zones are with me here or not but someone please help.
  8. Jun 22, 2006 #7


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    Well, ALL of them may contribute to the kinetic friction force!! So it is a very badly written question.

    The kinetic frinction force has a magnitude [itex] \mu_k N [/itex] where [itex] \mu [/itex] (Greek letter "mu", pronounced "mew") is the coefficient of kinetic friction and N is the magnitude of the normal force.

    Depending on the inclination of the surface on which the object and the direction of the applied force (if any), the magnitude of the normal force may or may not depend on the mass, the value of g or the applied force!

    EDIT: I just saw your drawing. In *that* case, the normal force is simply mg (the applied force does not affect the normal force) so in *that* case, the answer was "the applied force"
  9. Jun 22, 2006 #8
    Ok a few more about that one, the normal force is only dependent on mg in that one right? Like you said, then what it is the coefficient of friction dependent on? I believe I read about it a bit in my textbook, but I don't have it any more since my exam is done, what is it dependent on? Oh, and how do you write the 'mu' as you have, and the subscript? I read the thread about producing latex images, but I couldn't get it working...

    Also, how would this change if the object was on an incline as you mentioned?

    Oh, and I don't know if you already intended to or not, but would you be able to lend a hand with the other questions I still have, if it's not too much trouble. I appreciate the help you've provided thus far, and I hope you'll continue.
  10. Jun 22, 2006 #9


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    Yes. In fact , the normal force is equal to mg in that case
    It depends only on the *composition* of the two objects sliding against one another (steel on steel or ice on steel or whatever)
    Just go to my post and left click with your mouse on the formula, you will have a small window opening up that will show exactly the code I entered.
    It depends on how the applied force would be acting. If the applied force would still be parallel to the surface, the friction force would still not depend on the applied force (but the normal would be [itex] mg cos \theta[/itex] instead). However, as soon as the applied force is not entirely parallel to the surface (no matter if the plane is inclined or not), then the normal force will depend on the applied force (its magnitude and the angle at which it is applied).

    Sure, but I can't reread all the posts so maybe you could tell me which of the questions in your first post are still unanswered?
  11. Jun 23, 2006 #10
    Ok, so the only things affecting kinetic friction(on a horizontal plane) are the normal force, which is dependent on the force of gravity, so that can be included as well, right? Also, the [itex]\mu_k[/itex], the coefficient of kinetic friction, dependent on the materials in contact. The force of friction in the free-body-diagram was what exactly then? One of the factors going into the calculation of [itex] \mu_k N [/itex]?
    If the applied force was applied at an angle, then how would that affect the normal force? Would it increase in response to the added force into the ground(and thus providing a greater reaction force parallel to surface). And what about on an inclined plane? How is the normal force affected there, the normal force diminshes as the incline increases right?
    If all of this is right, then I can move on to the remaining ones.

    Now, about the other questions, the ones still remaining are as follows:

    Why does a car slip on ice exactly? If you push the tires backwards into the ice, it should still respond with a reaction force of equal magnitude, right? So what causes the slippping?

    Last, why do the molecules pull you in different directions in water, thus affecting gravity apparently?

    Umm, that's all right now I'm pretty sure, someone please help soon.
  12. Jun 23, 2006 #11


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    Right. As long as there is no other force with a component perpendicular to the surface. Basically, the normal force is such that [itex] \sum F_y =0[/itex]. If there is on ly gravity, you have N-mg=0 so N=mg. If any other force has a y component, the normal force will differ from mg.
    I am not sure what you mean. The kinetic friction force is not a factor in the calculation of [itex] \mu_k N [/itex], the kinetic friction force *is* [itex] \mu_k N [/itex]!

    Yes. (we are still talking about a horizontal plane I assume)
    It depends if the applied force has a y component which is negative or positive. In the first case, the applied force is partly "pushing the object into the ground" if you see what I mean. This means that the normal force will be greater than mg and therefore the friction force is larger than when the applied force is horizontal. But you could also have an applied force with a positive y component (think of a string attached to the object and someone pulling on the string at an angle above the horizontal). This will decrease the normal force and therefore decrease the friction force.
    Again it depends on all the forces in the problem so the question is too vague. Consider the special case of no applied force at all. There is just gravity. Then, as the angle of the incline increases, the component of gravity perpendicular to the surface decreases (it is [itex] mg cos \theta [/itex] with the angle measured with respect to the horizontal)). So you are right that in that case the normal force decreases as the angle increases.
    When tires rotate without slipping, the friction force is a static friction force. When the tires are slipping, the friction force is a kinetic friction force (which smaller than the static friction force). But there is still a friction force and there is a reaction force provided by the ground indeed.
    But if the surface is very icy, this kinetic friction force is very small.

    Imagine being on a perfectly levelled icy surface. You push on the gas pedal. The wheels start spinning. They apply a kinetic friction force on the ground (which is small because the surface is very icy) and the ground apply a reaction force on your car. You *will* start to move albeit with a very small acceleration because the forces are small.

    In real life, however, the surface is usually not perfectly levelled and the tires are in some grooves so the force is not enough to overcome gravity keeping you stuck in the grooves and you remain stuck.

    I am not sure what this means. Of course gravity is unchanged. What is affected is the "apparent weight" but the way this is defined, it is not the true weight which is still mg. The water molecules have random motion but the net force exerted by the water on you is not in "different" directions, there is a net force upward (the buoyant force). So I am a bit confused by this question.

  13. Jun 23, 2006 #12


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    Okay just relating your your question on friction, if we take a particle of mass m moving with a constant velocity with a force is being applied by a light inelastic string at an angle of [itex]\theta[/itex] degrees above the horizontal. The tension in the string is T. Now, as the particle is not acclerating the vector sum of all the forces should be zero. Now the tension will have a component in the horizontal plane ([itex]T\cos\theta[/itex]) and a component in the vertical plane ([itex]T\sin\theta[/itex]). Now, the friction force is given by [itex]F_{r} = \mu R[/itex], were [itex]\mu[/itex] is the co-efficient of friction and R is the normal reaction force. So if we now sum the forces;

    In the horizontal plane we have friction and the applied force;

    [tex]\sum\vec{F_{x}} = T\cos\theta - F_{r} = 0[/tex]

    And in the vertical plane we have gravity, the normal force and the applied force;

    [tex]\sum\vec{F_{y}} = R + T\sin\theta - mg = 0[/tex]

    Now rearranging the above equation to find a function of R in terms of [itex]\theta[/itex] we have;

    [tex]R = mg - T\sin\theta[/tex]

    Now substituting this into our equation for friction we have;

    [tex]F_{r} = \mu R = \mu(mg - T\sin\theta)[/tex]

    Can you see now why than angle at which the force is applied affects the normal reaction force and hence the frictional force? Observe that if the rope is horizontal [itex]\theta = 0[/itex] and the reaction force just becomes R = mg. Do you follow?

    EDIT: Looks like Patrick got there before me, I need to type faster :smile:
  14. Jun 23, 2006 #13


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    Perhaps, the example of water was a bad idea. This is slightly off topic, but I don't think it will do any harm (and I'm not doing anything till later:rolleyes: ). The real question is why does newton's law of gravitation break down at the surface of the earth. OKay, so let us take an object of mass m at a point inside the earths surface, a distance r from the centre of the earth thus;
    [PLAIN]http://eml.ou.edu/physics/module/newton/jik_leong/src_img/insideearth.jpg [Broken]
    Taken from the University of Oklahoma website

    An important point to note here is that the matter above the particle exerts no force what so ever on the particle, this is derived from the fact that a uniform shell exerts no gravitational force on an object placed within it. Another point to note is that the gavitational force experienced by the object is only due to the mass of the earth below it, i.e. inside the red ring in the above diagram. As the object moves closer to the centre of the earth, there are two antagonistic effects. The magnitude of the gravitaional force increases due to the reduced distance from the centre of mass (recall that the gravitational field stregnth is inversely proportional to the square of the distance between the two objects. However, the force also decreases due to the decreased mass enclosed by the red circle. If we assume that the earth is of a uniform density the latter is more significant than the former.

    As I said above, the mass of the particles above the object do not contribute to the net force and therefore can be ignored. Let the mass of the earth enclosed by the circle be M. The mass enlcosed by the red circle is given by the product of density ([itex]\rho[/itex]) and volume (v), assuming the earth is a sphere;

    [tex]M = \rho v = \rho\frac{4}{3}\pi r^{3}[/tex]

    Now if we substitute the above into newton's law of gravitation we obtain;

    [tex]F = -\frac{GMm}{r^{2}} = - \frac{Gm}{r^{2}}\cdot\rho\frac{4}{3}\pi r^{3} = -\frac{4\cdot Gm\rho\pi r}{3}[/tex]

    Now, if we compare the two equations, you will see that;

    [tex]\frac{4\cdot Gm\rho\pi r}{3} < \frac{GMm}{r^{2}}[/tex]

    Provided that r is less than the radius of the earth.
    Last edited by a moderator: May 2, 2017
  15. Jun 23, 2006 #14


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    I would not say that Newton's law of gravitation is breaking down. It is used in the rest of the calculation. but the equation [itex] G {m_1 m_2 \over r^2} [/itex] is only valid for point masses. What *is* breaking down here is treating the Earth as a point mass, not the law of gravitation itself (and it obviously breaks down here since one could not be inside a point mass:tongue2: ).

    Newton's law of gravitation does break down (for example in understanding the precession of the aphelion of the planets, most clearly detectable for Mercury) but not in this example.

    I know that Hoot knows all of that, I just wanted to clarify.


  16. Jun 23, 2006 #15
    Hoot, I still don't see why you say that Newton's law would break down. It is indeed universal . The shell theorem is basically a simple application, and if I remember correctly, it was Newton himself who proved it .
    I think the -ve sign is not required.


    Edit :Patrick's on a roll !
  17. Jun 23, 2006 #16
    I thought force was directly proportional to mass, wouldn't the force decrease due decreased mass?
  18. Jun 23, 2006 #17

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    Of course; I'm sure Hoot meant to write that "the force also decreases due to the decreased mass..."
  19. Jun 23, 2006 #18
    Ok, so how do I explain the original question then? Is the gravitational field strength strongest at the poles, or at the bottom of the ocean, and why exactly? Did I miss something here?
  20. Jun 23, 2006 #19

    Doc Al

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    What point is closest to the Earth's center? (What's the difference between equatorial and polar radii? How deep is the ocean?)
  21. Jun 23, 2006 #20
    I'm not sure... I never found out anything about that...
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