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Springs, Vectors, and a graph

  1. Oct 16, 2004 #1
    This is my first post...and I hope I am doing this correctly...I know this isn't a place to "get my homework done for me". I just want to know how to do it.

    3. [HRW6 7.P.022.] A 220 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.8 N/cm. The block becomes attached to the spring and compresses the spring 11 cm before momentarily stopping.

    (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) [in meters/second]

    (d) If the speed at impact is doubled, what is the maximum compression of the spring? [in meters]

    I don't know how to begin (c)....and (d) I assumed it would be either 4 times bigger or 2 times bigger than 11 cm -- which is wrong.

    (c) I started by using the equation....W = K-final - K-initial (W - work, K - kinetic energy) which is W = 0.5 * k * x-final ^ 2 ("k" is the spring constant). When I finished pluggin numbers in and solving I got the answer 0.86334 J yet this answer is wrong....what am I doing wrong?
  2. jcsd
  3. Oct 16, 2004 #2
    Springs, Vectors, and a graph Part 2

    Now the Vector problem.....

    7. [HRW6 7.PN.20.] A force F = (2.9 i + 7.5 j + 8.0 k) N acts on a 2.00 kg object that moves from an initial position of di = (3.0 i - 2.0 j + 5.0 k) m to a final position of df = (-5.0 i + 3.9 j + 7.1 k) m in 4.3 s

    (a) Find the work done on the object by the force in that time interval.

    (b) Find the average power due to the force during that time interval.

    (c) Find the angle between vectors di and df.

    Ok....I'm not to confident in my vector skills

    I'm not really sure how to do any of them....(atleast what equations need to be used)

    but to start (a) I found the change in d by subtracting d-final from d-initial which (if I did this correctly) gives (-8.0 i + 5.9 j + 2.1 k) m then I took that number found the magnitude of the line took the square root of (i^2 + j^2 + k^2) which yielded 10.15972 m then I did the same thing to the force...which is 11.3428 N then I multiplied them together W = F* d which is the number 115.2397 J which of course is wrong....I didn't include the angle (which I don't know how to do) and I think I messed up somewhere else.

    (b) and (c) don't know where to begin.....sorry...if you could give me a push in the right direction...I'd really appreciate it.
  4. Oct 16, 2004 #3
    Your first question deals with the conservation of energy. The kinetic energy of a particle is completly converted to the potential energy stored in a spring. Knowing that, you should be able to solve this.

    use:[tex]V_{spring}=\frac{1}{2}k(\Delta S)^2[/tex] where delta S is the change in spring state. Remember to convert your units to standard units (kg, m).

    I get somewhere in the ballpark of three m/s as my answer BTW.
  5. Oct 16, 2004 #4
    Finally the graph....

    I don't know how to add pictures.....so I'll explain the graph
    | | | | | | |
    |----------- 4
    |\| | | | | |
    |----------- 0
    | |\|_|_|_|_|
    |----------- -4
    | | | | | | |
    0 1 2 3 4 5

    ok, ok it is a crude graph.......now the explaination of it.....the y-axis is Force the x-axis is displacement.......

    the line starts at (0,4) then heads at a slope of -4 (y=-4x) untill hitting the (2,-4) [Note it crossed through the point (1,0)] then it travels with no slope to the point (5,-4).........that's the graph....hope that cleared it up :smile:

    now the questions:

    4. [HRW6 7.P.026.] The only force acting on a 2.4 kg body as it moves along the x axis varies as shown. The velocity of the body at x = 0 is 4.0 m/s.

    (a) What is the kinetic energy of the body at x = 3.0 m?

    (b) At what value of x will the body have a kinetic energy of 8.0 J?

    (c) What is the maximum kinetic energy attained by the body between x = 0 and x = 5.0 m?

    again I basically don't know where to start....I really don't know the equations to use either....

    I attempted (a). all I did was F * d [I know that is work and it is asking for kinetic energy] I got the answer....-12 J which of course is wrong.....please help
  6. Oct 16, 2004 #5
    For number 7,

    (a) Work is not F*d it is F*dcos(x) (where x is the angle in between them). However, you can't use this because you don't have the angle in between them. Instead, W is given as the dot product of F and d.

    (b) P(avg) = W/t

    (c) This question again involves the dot product. Can you see how?
  7. Oct 16, 2004 #6
    Thanks Parth Dave

    I just finished problem 7 and got all the right answers....I do appreciate your input...now on to the next problems :cool:

    and I understand it (which is always a plus)
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