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AP Physics questions Help

  1. Nov 7, 2003 #1
    Hi, I need help with my AP Physics questions. I tried all the problems and was unsure of my ways to approach the problems. If you can correct my errors, I would greatly appreciate your help. Thank you!!!!

    Please answer each question and tell me whether it is right and show me how to do it and your work.

    1. A 250.0g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2/5 N/cm. The block becomes attached to the spring and compresses the spring 12.0cm before temporarily stopping. While the spring is being compressed, what work is done on the block by.....
    a) the gravitational force on it?
    my answer)
    -> Work (gravitational) = F * d = (m*g)*h 0.25*9.8*0.12 = 0.29J (my question is that should the work have a negative value because the gravitational foce done on the block is downward?)

    b) the spring force on it?
    my answer)
    -> Work (spring) = 1/2(K*X^) = 1/2*250*0.12^ = -1.8J (What should the sign of the final value(1.8J) be? positive? negative?)

    c) What is the speed of the block just before it hits the spring? (Assume negligible friction and air resistance)
    my answer)
    -> W(gravitational) = delta K (change in kinetic energy) = 1/2(0.25)(vm/s)^2 - 1/2(0.25)(0m/s)^2
    v = 1.52 m/s

    d) If the speed at impact is doubled, what is the maximum compression of the spring?
    my answer)
    -> W(spring) - W(gravitational) = delta K (change in kinetic energy)
    That is all I could think of. I don't know how else to try this question. I have a test tomorrow, and it will be on this kind of questions. Please help me....

    2. A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200.0kg, which is required to travel upward 54.0m in 3.0minutes, starting and ending at rest. The elevator's counterweight has a mass of only 950.0kg, so the elevator motor must help pull the cab upward.
    a) What is the total work done by all forces on the elevator - counterweight system?
    my answer)
    -> Work = delta K (change in kinetic energy) = 1/2(1200kg)(final velocity)^2 = 1/2(1200kg)(om/s) = 600(final velocity)^2 - 0
    (final velocity) = 54m / 3min = 0.3m/sec
    Work = 600kg(0.3)^2 = 54J (so I got total work equals 54J. But the question asks for the total work done by all forces on the elevator and counterweight system. I don't understand how to find counterweight system. Does my answer include everything that the question asks for? or what should I do?)

    b) What is the work done by the motor on the system?
    my answer)
    -> I really can't do this problem until I finish the part(a). But my idea is that I should subtract the gravitational work from the work of the system. Am I right? If so, I just do mgh(cos180) to find the value of gravitational work. Right?)Please answer each question and tell me whether it is right and show me how to do it and your work.
  2. jcsd
  3. Nov 7, 2003 #2


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    Something's missing here: what is the distance between the initial height and the top of the relaxed spring? Is that 12 cm too so that it falls a total of 24 cm? And is k = 2.5 N/cm?

    As far as the sign on work goes: in problems like these, if the force doing the work points in the same direction as the displacement of the object, it is positive work; if they point in opposite dierections, it is negative work.
  4. Nov 7, 2003 #3
  5. Nov 8, 2003 #4


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    I'm not sure about the sign on the work. It's a bit tricky because it's conversion from kinetic to potential energy. I suppose a good guideline would be to consdier anything that slows the block down negative, and anything that speeds it up as positive.

    Ok, so here's the approach I would take to this problem:

    The energy that ends up in the spring is 1/2 kx2.
    Since the only way that the block is gaining energy is by gravity, the work done by gravity is equal to the energy in the spring.

    So, K=0.4 N/cm (2/5=0.4)

    Now, the total work is 1/2 kx2
    so you get:
    .5 * .4 N/cm * 12cm * 12cm=28.8 Ncm
    (Check my math)
    as the total potential energy = the total work done by gravity.

    b) Spring force = kx where k is spring constant, x is displacement. This should be *really* easy.

    c) This one is a bit tricky:
    There are two things that are working on the block while it is on the sping: Gravity, and the Spring. If you account for both, then you can calculate the kinetic energy when the block hit the spring.

    d)This is pretty straightforward if you get part c, but remeber that gravity is doing work on the block while it's going down the spring.

    If you know calculus, c and d may be easier if you do work integrals based on net force instead of dividing the work into components for the spring and for gravity.

    Question 2:
    Part a:
    The elements in the system are the counterweight and the elevator. You can assume that the elevator's change in height is the opposite of the block's change in height. Since both start and end at rest, you can calculate change in potential energy to get the work.
    (BTW, I think you transcribed this incorrectly, the question probably asks for the total work done on the elevator.)
    Part b:
    see part a since all of the work is due to the motor.
  6. Nov 8, 2003 #5

    Doc Al

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    Since the force (the weight) is in the same direction as the displacement, the work done by gravity is positive.

    The spring pushes up, but the block moves down: work is negative.

    The block hits the spring with a certain KE (and thus speed). The work done on the block (by gravity and the spring) must equal that initial energy (at the point that the block stops).

    You have the new KE (if the speed doubles, the KE is 4 times as much). Now the net work by gravity and the spring must be 4 times as much. Calculate the new displacement.

    The total work done by gravity and the motor (ignore friction, etc) equals zero, since it begins and ends at rest. The time is a red herring.

    See above. To calculate the work done by gravity, realize that: When the cab goes up, the counterweight goes down. The net work by gravity is negative, meaning that the motor must add some work to move the cab.
  7. Nov 8, 2003 #6


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    Obviously what I said about the height was wrong. Sorry about that, but there's no need to shout; I was only trying to help. I'll try not to repeat what others have said too much:

    1. a.) Set your datum as the lowest point the mass reaches. No energy is lost in the system, so the total energy of the system is given by .5*k*x^2= .5*(250)*(.12)^2 = 1.8 J. The net work done on the object (from its release point at the top to where it temporarily stops on the spring) is 0. The work done by gravity is positive and equal to 1.8 J (You can use this to find the initial height H by setting mgH = 1.8J, where (H-12cm) is the height the block starts out above the relaxed spring).

    b.) There are only 2 forces acting on this object: 1 due to gravity and 1 due to the spring. We already said the net work = 0 and gravity does +1.8J. The spring, therefore, does -1.8J of work on the block (=-.5kx*2)

    c.) Use conservation of energy: just before it hits the spring, the energy of the block is given by .5*m*v^2 + mg(12cm). This energy must be equal to the energy it has at the bottom of the spring (.5*k*x^2) = 1.8 J (since we assume no frictional losses in this problem). You should check the math yourself, but I get 3.47 m/s as the velocity.

    d.) You'll end up with a quadratic in x to solve for here. Let V = 2v = 6.9m/s, the new velocity of the block at impact. We'll redefine the datum as the lowest point the block reaches, which is no longer 12 cm of spring compression, it is the unknown. This means the conservation of energy will give us:
    .5*m*V^2 + mgX = .5*k*X^2

    Solve for X, the maximum compression of the spring (I just used a new variable in an attempt to avoid ambiguity with our previous x.)

    2. I think there's enough in the thread already for you to solve this one.
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