PHYSICS(help)

A small block of mass m=1 kg glides down (without friction) a circular track of radius R=2 m, starting from rest at height R. At the bottom of the track it hits a massless relaxed spring with spring constant k= 11 N/m, which starts to be compressed as the block continues to move horizontally. Note that we assume no energy loss during this “collision". There is friction between the block and the horizontal surface, and it is not uniform. As a function of distance, the friction coefficient varies like μ(x)=αx, with α= 1.0 m−1. Assume for simplicity that static and dynamic friction coefficients are the same, and use g=10 m/s2. (See figure)


(a) What is the maximal distance x1 that the block moves horizontally away from the track at x=0? (in meters)

x1=

unanswered
(b) What time t1 does it take for the block to travel between x=0 (relaxed spring) and x=x1 (block at first stop)? (in seconds)

t1=

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  1. I assume that you found the speed of the block as it left the track and hit the spring from conservation of energy
    Call it Vi = sqrt(2gh)= sqrt(2gR)
    the Ke of the block when it hits the spring is then
    Ke = (1/2)m Vi^2

    the retarding force of friction is
    Ff = m g mu
    where mu = 1x
    so Ff = m g 1 x
    the retarding force of the spring is
    Fs = k x
    in compressing distance d the total work done against friction and spring is
    Integral 0 to d of (mg+k)x dx
    which is
    (1/2)(mg+k)d^2
    so in the end
    (1/2)m Vi^2 = (1/2)(mg+k)d^2

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    posted by Damon
  2. Hi Damon, so we have to find 'd' from this equation? and how to find time?

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    posted by asha
  3. @Damon: thx
    @asha:
    solve for d:
    (1/2)m Vi^2 = (1/2)(mg+k)d^2
    -> d=sqrt(m Vi^2/(mg*alpha+k))

    but I also don't know the time

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  4. "d=sqrt(m Vi^2/(mg*alpha+k))"

    whats that alpha in your solution?
    shouldn't it be
    d=sqrt(m Vi^2/(mg*+k))?

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    posted by Greco
  5. Think of it as a harmonic oscillaton, and then find k', which does not equal k.
    After that, you can find the period( T) of the motion, what's the relation between t(x_0->x_1) and T.

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    posted by BBB
  6. I used T=2*pi*sqrt(m/k) and I divided it by 2, but it was wrong. Please BBB can you tell more about it?

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    posted by asha
  7. I also tried this to find k':
    (1) when spring is compressed
    1/2mv^2=µmgx+1/2kx^2
    (2) when spring returns back
    1/2kx^2=µmgx
    hence from (1) & (2) :
    1/2mv^2=2µmgx
    x=(1/4)* v^2/µg
    and
    1/2kx=µmg
    k=2µmg/x

    since µ=alpha*x I used µ=alpha*x1 (x1 found in part a)

    I then used t1= half a period
    t1 = pi*sqrt(m/k')

    But it's wrong.
    I lost the game :) but I want to understand what I did wrong and what is the good way to do it.

    I also had c) wrong... based on my not good at all b) :)

    Thank for the explanation!

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  8. did someone how to do 3,5,6b....?

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    posted by A
  9. I got a and c but the time is breaking my brain in 2. Any ideas?

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    posted by Daoine
  10. Give me your c answer and I let you know my b correct answer

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  11. c) The block will stay put forever at x=x1

    for b) I get it's something with the W-E theorem but......friction being a non concervative force, energy is changed into heat.

    b)I just....don't trust myself anymore...
    loss of self confidence is a killer, you doubt things you'd never doubted otherwise

    thanks!

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  12. t = pi / (2*w) where w = sqrt((k+alpha*m*g)/m)

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  13. What is alpha?

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  14. NVM

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  15. (b) Asha - why did you divide by 2? Think about what a "period" of oscillation is. Leave the physics and maths to one side for a second. How is a "period" defined? Visualise it.

    For our motion from x to x1, how much of a period is that?

    (c) I got a different result, but this could be due to your values.

    Break this down into 2 parts.

    First part - does the block move back at all? This depends on whether the friction at x1 is greater than or less than the spring force at x1.

    You have expressions for both the spring force and the frictional force already, just plug in the numbers.

    If frictional force is greater than spring force, the block stays put.

    If not, move onto second part - energy considerations.

    How much mechanical energy do we have at x1? Easy to calculate.

    How much energy is lost due to friction on the return journey? You've already worked this out for part (a).

    So knowing whether the mechanical energy is greater or less than the energy lost to friction, you can deduce how far back the block moves.

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    posted by Pete
  16. check in the problem alpha(á) is "ì(x)=áx, with á= 1.0 m−1" or whatever value you have.

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  17. for:
    k=11 N/m
    a = 1.0
    m = 1 kg
    g = 10 m/s^2

    the equation is:
    t = pi/(2*sqrt((11+1.0*1*10)/1))) = 0.34277 sec

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  18. @ Pete
    if spring force and friction force are the same won't the masss stay put at x1?

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    posted by some0ne
  19. @Pete
    c) How do we deduce how far back the block moves with te energy difference obtained?

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    posted by Roy
  20. q6)a)x_1=sqrt(2*g*R*m/(m*g*alpha+k))
    b) t=pi/(2*sqrt((k+alpha*m*g)/m)))
    c) The block will move back and get catapulted up the circular track

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    posted by KS
  21. To answer part a)
    1. use conservation of energy and obtain how much energy is stored in the block when it reaches the bottom, obviously it is "mgR" since the height is the radius of the ramp.

    2. Now this object will transfer energy to the spring and friction will also convert some of the energy to heat.

    3. To know the expression of the energy transfered to the spring system, first find the forces acting on the block once it reaches the spring system.

    F = -kx - Nmu
    F = -kx -mg*(alpha*x)

    now that you have the expression use the physical definition of potential energy "V"

    -dV/dx = F hence
    dV = -Fdx

    integrate (kx+mg*alpha*x)dx

    now you'll get

    1/2kx^2 + 1/2mg*alphax^2

    equate that to mgR and you will obtain an expression to calculate for x1

    sqrt(2mgR/(k+mg*alpha)) = x1

    Now for part b)

    Remember that from x0 to x1 only represent 1/4 the period hence getting an expression for the period will solve this problem.

    Now remember that F = kx+mg*alpha*x
    and
    ma = (k+mg*alpha)x

    hence a = [(k+mg*alpha)/m]x

    that is pretty much like w^2

    w = sqrt((k+mg*alpha)/m)

    2pi/T = sqrt((k+mg*alpha)/m)

    hence

    T = 2pi*sqrt(m/[k+mg*alpha])

    and since w said that tx0->x1 is T/4

    now we have

    t = (pi/2)*sqrt(m/[k+mg*alpha])

    For the third part, it highly depends on the value of x1

    First compare mg*alpha*x1
    is it larger or smaller than k*x1

    if k*x1 = mg*alpha*x1
    or k*x1 < mg*alpha*x1

    then the block would stay at x1

    but if k*x1 > mg*alpha*x1

    then the block would either stay between x0 and x1 or move back to the circular track.

    To know which would happen check the potential energy of the block at x1

    then compare it to the work done by friction.

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  22. @some0ne

    If the spring force and the friction force are equal and opposite, what is the net force?

    @Roy

    "How do we deduce how far back the block moves with the energy difference obtained?"

    The friction is a function of distance. On the way from x to x1, a certain amount of the initial mechanical energy is used up, until the block reaches x1 and has compressed the spring by x1.

    *IF* you conclude from part (b) that the block moves back (and doesn't remain at rest), then you have to figure out what happens on the way back.

    On the way back from x1 to x, the friction is still a function of distance, only this time it starts greater and decreases. That doesn't really matter. If the block were to travel all the way from x1 to x, then it would lose the same amount of energy through friction as it lost on the way from x to x1.

    You therefore know how much spring energy it *needs* to get back to x. Just compare the spring energy it actually has, with what it needs, and the answer will jump out at you.

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    posted by Pete

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