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Physics Question Thread 2

Discussion in 'MCAT Study Question Q&A' started by Pembleton, 07.18.04.

  1. Pembleton

    Pembleton Senior Member

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    Hi guys,

    I don't have my physics textbook nearby and I was wondering if someone could remind me how to figure out the center of mass of two objects on a plank, for instance.

    For example a 10 kg box is 1 m from the edge of a 10 m plank and a 20 kg box is 2 m from the other edge of the plank. Where is the center of mass on the plank?
  2. MDtoBe777

    MDtoBe777 Senior Member

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    First of all, you need to know if it is a masless plank or not.

    10 kg box is 1 m from the (left) edge of a 10 m plank and a 20 kg box is 2 m from the other edge (right) of the plank. Where is the center of mass on the plank?

    Call the left side of the plank 0. Then the 10 kg box is at position 1 m.
    Call the right side of the plank 10m. Then the 20 kg box is at position 8m.

    Xcm = (M1X1 + M2X2) / (M1 + M2)

    Now plug it in.

    Xcm= [ (10)(1) + (20)(8) ] / (10 +20)
    10 + 160 /30
    = 170/30 (or about 17/3 m roughly 6. something) ? This answer would be different is the rod was not masless.

    Please correct me if I'm wrong.
  3. willthatsall

    willthatsall Unretired

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    I agree with the poster of ambiguous gender. :D :thumbup:
  4. MDtoBe777

    MDtoBe777 Senior Member

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    edit: her..haha
  5. Spartacus

    Spartacus X & Y

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    yeah, mdtobe is right
  6. QofQuimica

    QofQuimica Seriously, dude, I think you're overreacting.... Administrator SDN Senior Moderator Lifetime Donor

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    This thread is a continuation of the original Physics Question Thread. Feel free to post your physics questions here. :)

    If you are new to the MCAT subforum, please see the Physics Question Thread rules here.
  7. gridiron

    gridiron Moderator Emeritus

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    Thanks for opening the new thread!

    Question for users: Are there any topics or content you would like to see covered in the Physics FAQs and Topic Writeups thread? This can be anything from MCAT physics questions to physics content. If you do, please post here! Thanks!
  8. dill_punjabi

    dill_punjabi

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    Does an object near earth surface experience both gravity and universal gravitational force?
  9. gridiron

    gridiron Moderator Emeritus

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    Yes, an object near the surface of earth will experience both forces. Both forces actually do the same thing and mean the same thing. Gravity, from the perspective of earth, is the force of attraction of earth to some body near its surface. This is denoted by weight which equals mg. The gravitational force also explains the mutual attraction of earth to an object near its surface. I hope this helps and good :luck:.
  10. Comdessert

    Comdessert

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    Hi there,

    I'm having trouble getting my head wrapped around this one.

    A 50kg woman dangles a 50kg mass from the end of a rope. If she stands on a frictionless surface and hangs the mass over a cliff with a pulley, what is the tension on the rope (this is from EK's 1001 questions and includes a diagram)?

    The solution says 250N, but the reasoning provided is opaque to me. If someone could lead me through this I'd really appreciate it.

    Thanks,

    - CD
  11. BrokenGlass

    BrokenGlass

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    Let m1 represent the 50kg mass and m2 represent the woman. Choose the mass and the woman as our system. So tension is an internal force to our system. The only outside forces acting on our system are weight of the woman, the normal force on the woman, and the weight of the 50kg mass. The weight of the woman and the normal force on the women cancel each other out. The NET force acting on the system is just the weight of the 50kg mass.

    net force = m * a (Newton's 2nd law)

    m1*g = (m1 + m2)*a

    50 * 10 = (50+50) * a

    a = 5

    Now choose the 50kg mass as our system. The only forces acting on this mass are its weight and the tension in the string.

    net force = m * a (Newton's 2nd law)

    m1* g - T = m1* a

    50 * 10 - T = 50 * 5

    500 - T = 250

    T = 250N
  12. wes431

    wes431

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    I have a question from 1001 EK. The diagram shows a pulley with two masses hanging on both sides. On the left side is mass of "m" and on the right side is mass of "2m".

    The question says:

    The masses below hang across a massless, frictionless pulley. What is the tension in the rope.
    a) 0.5 mg
    b) mg
    c) 1.33 mg
    d) 2 mg

    I keep getting 0.33 mg but it says the answer is C. Is this just a mistake in the book?
  13. Foghorn

    Foghorn Guest

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    Look up Atwood's machine. You're probably just putting in the wrong sign (+/-) for the acceleration on one of the masses.

    Eq 1: Force on mass 1 = T - (mass 1)*g = (mass 1) * a

    Eq 2: Force on mass 2 = T - (mass 2)*g = (mass 2) * (-a)

    Solving for a by subtracting Eq 2 from Eq 1 : [(mass 2 - mass 1)/(mass 1 + mass 2)]*g

    Solving for T: [(2*mass 1*mass 2)/(mass 1 + mass 2)]*g
  14. BrokenGlass

    BrokenGlass

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    Obviously the masses will accelerate so that the smaller mass goes up and the larger mass goes down. If tension was 2mg, then larger mass would be in equilibrium, but it's accelerating down, so T < 2mg. If tension was mg, then the smaller mass would be in equilibrium, but it's accelearting up, so T > mg. The only answer choice between mg and 2mg is 1.33 mg
  15. wes431

    wes431

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    Well I tried doing it mathematically, but I don't know what I did wrong. Let me know please.

    for the "m" mass
    T-mg = ma

    for the "2m" mass
    T-2mg = -2ma or T = 2mg-2ma

    substituted the second equation into the first
    2mg-2ma-mg = ma
    mg = 3ma
    a = 1/3g

    therefore, T = 1/3mg
  16. BrokenGlass

    BrokenGlass

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    After you find acceleration, which is g/3, there is one more step. Choose mass m as your system. Apply Newton's 2nd law:

    T - mg = ma
    T - mg = m*(g/3)
    T = mg + m*(g/3) = mg(1 + 1/3) = (4/3)mg = 1.33 mg
  17. Comdessert

    Comdessert

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    Hey BrokenGlass,

    Thank you, that makes a lot more sense now! Looks like I'm going to need to practice figuring out systems. :)

    - CD
  18. wes431

    wes431

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    Thanks Brokenglass, but one more question. Why do you choose mass "m" as your system?
  19. wes431

    wes431

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    Another question.

    A spring (400N/m) is cut in half to make two new springs. What is the spring constant of each of the new springs?

    A. 100 N/m
    B. 200 N/m
    C. 400 N/m
    D. 800 N/m (answer)
  20. gridiron

    gridiron Moderator Emeritus

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    Hey! The shorter the spring, the stiffer it is and thus the larger the spring constant it will have. This means that the spring constant of a spring is inversely proportional to the length of the spring. From the question above, if you cut the spring in half the spring constant will double for the two new springs. I hope this helps and good :luck:.
  21. BrokenGlass

    BrokenGlass

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    The spring constant is a measure of how stiff the spring is. A spring that is very hard to stretch out has a large spring constant.
    A spring that is easy to stretch has a small spring constant.

    Here is a similar problem.

    Example: Suppose a 10-coil spring has a spring constant k. If the spring is cut in half, into two 5 coil springs, what is the spring constant of each of the smaller springs?

    Solution: Suppose we apply a force that compresses a 10-coil spring by 1 cm, then each coil is compressed by 0.1 cm. Now compress a 5-coil spring by 1 cm, then each coil will be compressed by 0.2 cm. According to the ideal spring equation the force of compression for a single coil is directly proportional to the displacement of the coil. The force needed to compress a single coil by 0.2 cm must be twice as large as the force needed for a 0.1 cm compression. Thus, the spring constant of the 5-coil spring must be twice that of the 10-coil spring. In general, the spring constant is inversely proportional to the number of coils in a spring so shorter springs are stiffer springs.
  22. BrokenGlass

    BrokenGlass

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    This choice is somewhat arbitrary. You could choose the other mass as your system in step 2. We are trying to solve for tension, so either mass can be chosen as our system in step 2, because tension is the same for both masses.

    Personally, I prefer the method that doesn't involve math. Math is always error-prone.
  23. BrokenGlass

    BrokenGlass

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    And I am going to need to practice figuring out women
  24. wes431

    wes431

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    This is problem 256 in EK 1001 if you have it. The diagram shows a spring placed horizontally. It is compressed 5 cm at each end of the spring with a 50 N force.

    The question asks, What is the spring constant k for the spring shown below?

    A) 0 N/cm
    B) 5 N/cm (answer)
    C) 10 N/cm
    D) 20 N/cm

    I put C as the answer because I imagined placing the spring vertically and placing a mass on it causing a force of 50 N and displacement of 5 cm. Therefore, I thought it would be (50N)/(5cm) = 10 N/cm. I kind of understand why it's C, but not really sure on the explanation.
  25. BrokenGlass

    BrokenGlass

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    x = 10cm, not 5cm because the spring is compressed by 5cm at each end. There is no need to imagine the spring vertically. In fact it's placed horizontally so that force due to gravity doesn't enter into consideration.
  26. wes431

    wes431

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    Alright so I did understand that but this other problem right above it reallly confused me.

    A man holds one end of a 30 cm spring in each hand. If he applies a 100 N force to each end of the spring, by how much does he shorten the spring?
    (k = 1000 N/m)
    A) 5 cm
    B) 10 cm
    C) 20 cm
    D) 30 cm

    According to what you just said, shouldn't it be C. But the answer in the back is B. What's the difference between these two probs?

    And Brokenglass, thank you so much for answering my questions. How much do you charge? :)
  27. BrokenGlass

    BrokenGlass

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    F = k*x (we can ignore the negative sign here for the purposes of this problem).

    x = F/k = (100N) / (1000 N/m) = 0.1 m = 10 cm. Why do you think it should be C?

    Each end is compressed by 5 cm, since the total amount of compression is 10 cm.
  28. wes431

    wes431

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    Why is each end compressed up 5 cm. Shouldn't each end be compressed by 10 cm. Thus making the total compression 20 cm.
    x = (100 N)/(1000 N/m) = 0.1 m = 10 cm just for one end
  29. BrokenGlass

    BrokenGlass

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    No, x represents the TOTAL displacement (stretching or compressing) from the equilibrium position.
  30. wes431

    wes431

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    Which of the following would be true concerning a spacecraft and the moon when they are in the same orbit? (Assume that netiher is using a propulsion system to maintain its orbit)

    A. They both must be at the same speed (answer)
    B. They bost must have the same mass
    C. They both must have the same mass and speed
    D. They must have different masses
  31. BrokenGlass

    BrokenGlass

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    centripetal force = mv^2/r (remember centripetal force is the net force)

    GMm/r^2 = mv^2/r (now cancel m from both sides)

    GM/r^2 = v^2/r (now multiply both sides by r)
    GM/r = v^2

    v = sqrt (GM/r), where sqrt stands for square root

    Being in same orbit means r is the same for a spacecraft and the moon. So yes, they will both have the same speed. The mass is irrelevant since it drops out.
  32. gridiron

    gridiron Moderator Emeritus

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    Another way you can solve the problem is by understanding the relationship between speed and radius of orbit. From kinematics, D = RT, where D is distance, R is rate and T is time. To find the speed you would need to divide the distance travelled by the time. For a circular orbit, the distance travelled is given by the circumference of the circle--this is 2*pi*R. Thus, speed would be equal to: 2*pi*R/T. For two objects, regardless of mass, travelling in a circle of the same radius, they will have the same speed.
  33. coolioyo

    coolioyo New Member

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    I got this question from a Kaplan Full length?

    What happens to the velocity of water pouring from a hole in a bucket of water when it is moved to a higher altitute.

    The formula is square root of 2gh. So, I though g changes and thus v. But they mentioned nothing about that and say velocity is constant.

    Also what happens to the water level if a block of ice floating in a cup of water melts, does it rise or descend and why? Thank you
  34. gridiron

    gridiron Moderator Emeritus

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    Hey! For the first question, can you post the full question and choices? To answer your second question--the water level will not change. When water freezes, it expands. In ice, water molecules are more spread out than in water. Since the water molecules are more spread out in ice, the density of ice is less compared to the density of water. Because the solid form is less dense than the liquid form, this is why water is weird, ice will float in water and when it is placed in water a certain amount of water will be displaced--this depends on the mass of the ice. However, when the ice melts, all the water will fill back up. The ice will shrink as it melts. To understand this better, try an experiment. Fill a glass to the rim with water and place some ice cubes inside. What will happen? The water level will rise. Let the ice melt and you will see that water will fill back up into the glass--the amount that was displaced by the ice cubes.
  35. knightstale4

    knightstale4

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    Three questions, and thanks in advance to whoever answers them! :D
    1. I was reading the NOVA book, and it said that the normal force is not always equal to mg or gravitational force. In what instances would this be true? Some examples please...
    2. How do you determine the direction of torque? Is it simply clocks are negative, and a positive value= counterclockwise, and negative value = clockwise direction?
    3. Is there a simpler way to understand right hand rule #1?:( It says your fingers should point in direction of B, and your thumb in the direction of velocity? What if angle between v and B is 150? Then which direction would your thumb be pointing? And are there only 4 directions for the force: up, down, into the page, and out of the page?
  36. majik1213

    majik1213

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    1.

    -Imagine a block such that it rests on a flat surface while a rope provides some vertically upward tension (i.e. the rope is affixed to suspend the block somewhat). Here, the normal force is equal to the gravitational force minus the tension in the rope.

    -In outer space, free of the force of gravity, imagine a collision between two asteroids. When the two objects collide, they exert a normal force on one another.

    -Imagine two magnets with a wooden block between them in outer space, free of any gravitational forces. The magnetic force causes the two magnets to attract each other, and four normal forces arise. Each side of the wooden block exerts a normal force against the magnet that is equal in magnitude but opposite in direction of the magnitude of the magnetic force that each magnet exerts on one another, adding for a total of two normal forces on the magnet due to the wooden block. In addition, each magnet exerts a normal force on the block, for a total of two additional normal forces on the wooden block due to each magnetic face that is in contact with the block (two faces total, one from each magnet).


    I don't know your understanding about the normal force in general, but let me say this: The normal force is actually an electrostatic force. It comes from the Pauli exclusion principle, which disallows electrons with parallel spins to fill the same orbital levels. In other words, the normal force has nothing whatsoever to do with gravitational force per se, but the force of gravity can trigger the occurrance of the normal force.

    2.
    Torque is a vector valued definition. Torque equals r cross F, where r is the radius from the chosen origin (the point about which angular motion occurs), and F is the force on the object. Using the right hand rule, we can find the direction of the torque. Your guesses of clockwise or counterclockwise are non-sensical. Torque has no simple direction; you must always find it with the right hand rule (R.H.R.).

    3.

    Here is my version of the right hand rule. If you are crossing two vectors, A cross B, point your index find in the direction of A and your middle finger should bend until it is aligned with the direction of B. (WARNING: You MUST ensure that the two vectors, A and B, are joined at the tail before doing this procedure.) Then your thumb will naturally point in the direction of the vector that represents A cross B.

    if the angle between v and B is 150 (degrees presumably), then just follow the aforementioned rule.

    There are 6 basic directions from the origin in cartesian coordinates: +x, -x, +y, -y, +z, and -z. When you said, "up, down, into the page, and out of the page" you forgot left and right. Any direction can be a combination of these directions as well.
  37. majik1213

    majik1213

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    the velocity will change according to the acceleration of the water due to the gravitational force. Most likely, however, the answer is the velocity is constant because the change in height is negligible compared to the radius of the earth; therefore, you may be told that the velocity doesn't change. In fact it does, but the change is tiny.

    Well it looks like I was right about what they would assume.

    I am going to guess rise. The reason is that your system before the melting occurs is x% liquid and y% solid (and x+y=100%, for simplicity). This system's state contrasts the state it ends up in after the melting occurs, in which case now the system is 100% liquid and 0% solid. Because x<100%, there is more liquid now than before; therefore, we'd expect to see the level of liquid rise. In this case, the liquid is water.

    I might be wrong, because I know water expands as it freezes and contracts upon melting. But to be wrong, I need more information about what the question is asking or how the water level is being measured. Are you in a closed container, or even in a container at all?
  38. majik1213

    majik1213

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    I don't know if that answers the question s/he is asking very well because you must conclude somehow that the time taken for the very same distance to be travelled is the same for both objects. But if you know that, then you probably got the answer right because it is very easy to conclude that, because D=RT, R must be equal for both objects. The poster, however, presumably got the answer wrong, so s/he likely did not know that D and T were equal for both objects.
  39. BrokenGlass

    BrokenGlass

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    You are definitely a Physics major.
  40. Comdessert

    Comdessert

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    Hey folks,

    I've got two more questions, both from EK's 1001 Physics books.

    (Question 358 - includes a figure)

    There is a frictionless curved surface (a quarterpipe) with a 500 gram block resting on it. The block is 4 cm off the ground. The block starts from rest and slides down the curve. At the bottom of the curve, there is a flat portion with a kinetic friction coefficient between the block and flat area of 0.2. How far does the block travel along the flat portion of the ramp?

    (Question 367)

    In 0.5 seconds, a hammer drives a 30 cm nail into a piece of wood. If the friction force between the nail and the wood is 200 N, approximately how much work is done by the hammer?

    Thank you!

    - CD
  41. gridiron

    gridiron Moderator Emeritus

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    You are correct from a physics perspective. However, for MCAT purposes, objects traversing a circle in the same period will have different speeds depending on the distance from the center of the circle---the further the radius, the faster the speed.
  42. axp107

    axp107 UCLA 09': Italian Pryde

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    1) set mgh = 0.5 m v^2

    calculate v ....

    now, its encountering a frictional force opposing it.. so to calculate the frictional force, do

    Normal force * coefficient of kinetic friction and set it equal to ma..

    so,

    mg(0.2) = ma

    Find a, the acceleration to find the acceleration (or deceleration) that the object is slowing down at.

    At this point, don't worry about the sign of the acceleration.. (its in the opposite direction, but don't worry)

    So now we have the initial velocity, V0 at the bottom and the acceleration that it's slowing down at, a.

    use the Kinematics formula... V(final)^2 = V0^2 - 2a(d)

    Set V final = to 0 to find the V where it stops. Signs don't matter at all at this point and you get V0^2 = 2(a)(d)

    ***At first I thought you couldn't simply plug in acceleration since its slowing down! EXCEPT, setting V final = to 0 takes care of everything and you don't have to worry about anything ***

    Substitude and solve for d!


    2) I have no idea how to do #2 numerically... if its a multiple choice Q, all I can say is that

    its more than 200 N * 0.3 m..... W = F * d

    since the force of the hammer is more than friction... otherwise the nail wouldn't go in.
  43. wes431

    wes431

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    This is from EK 1001 Questions, problem 295. There's a drawing showing a F vector put on top of the box 60 degrees to the horizontal. The box weighs 5 kg.

    The coefficient of kinetic friction between the block and the surface upon which it rests is 0.1. The force F is 100 N. At approximately what rate does the block accelerate?

    A. 1 m/s2
    B. 4.5 m/s2
    C. 6 m/s2
    D. 7.3 m/s2
  44. BrokenGlass

    BrokenGlass

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    Is vector F pulling the box up or pressing it down at 60 degrees to the horizontal?
  45. wes431

    wes431

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    pressing it down
  46. xlr8

    xlr8 New Member

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    I was reading a question above:

    "Also what happens to the water level if a block of ice floating in a cup of water melts, does it rise or descend and why? Thank you"

    Someone answered by saying it would rise, but using a different line of reasoning I find it to descend. Since the ice is floating, that means its density is not greater than that of water, less than 1. Since the mass of the ice remains constant as it melts to water,
    Volume (of melted ice) = mass / 1
    Volume (of ice) = mass / (<1),
    so Volume (ice) > Volume (melted ice), and the ice will take up more space than its melted counterpart

    Am I wrong? Or does it ascend?
  47. free_radical

    free_radical

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    because the ice is less dense, not all of it will be submerged, so the water level will not be displaced in an amount equal to volume of the ice cube.

    so the issue is, even though the mass is constant and ice is more voluminous, maybe the melted ice displaces more water than the ice itself. or maybe it displaces less, because the lack of complete submersion doesnt compensate for the increased density of the melted ice (water).

    assume the ice has a SG of .8, so 80% of the volume was submerged, but then the ice cube is 1.25 times as voluminous as the melted ice.

    so premelt: 80% of 1.25 is in the water. post-melt: 100% of 1 is in the water. they're the same thing.

    so doesnt the water level stay the same? have i gone horribly off-track with this one?
  48. BrokenGlass

    BrokenGlass

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    You have weight of the box (50 N) acting down. You have Y component of F acting down (100 * sin 60 = 87 N), you have X component of F acting horizontally (100 * cos 60 = 50 N). You have normal force acting up.

    Normal force = 50 N + 87 N = 137 N (the box is not accelerating in the Y direction, so the NET force in the Y direction is zero).

    f kinetic = mew kinetic * N = 0.1 * 137 = 13.7 N.

    F net = 50 N - 13.7 N = 36.3 N.

    F net = ma => a = F/m = 36.3/5 = 7 m/s/s

    Unless I am misunderstanding the problem here, which is why when people ask questions, they should provide the answer key and explanation if possible.
  49. J DUB

    J DUB Watch my TAN walk!! Lifetime Donor

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    I will try to answer the second question.

    First off, you have kinetic energy taking place when you drive the nail and it ends when you finish driving the nail in.

    Therefore, you want the KE final to be zero.

    So, KE final = KE initial - your Work done by friction.

    You can find W friction = Fd = (200N)(.03m) = 6 J

    Plug the back into KE final = KE initial - Wfriction

    0 = KE initial - 6J, solve for KE initial

    KE initial = 6J So, the hammer had to do 6 J of work.
  50. tennisboy85

    tennisboy85

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    Can someone check if my reasoning is right? This topic is on bouyant force.

    Why does a piece of bread float in water, but when you crumple it into a ball, it would sink in water?

    Because isn't Fb = density (fluid) * volume submerged * gravity

    Of these, density and gravity can't be changed. So, is the reason because volume = surface area * height? Therefore, the greater the surface area, the higher Fb is. So conclusion = since weight of bread is same in both cases, Fb is higher in the "flat" bread over "crumpled" bread, that's why it floats?

    On similar note: If that argument was true, does this mean that sheets of metals would float if you spread it thin enough? Despite the fact that density of metals are much bigger than water?

    Thanks guys.

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