Finding the spring constant and the distance of the block?

the question is too long, i couldn't write it or upload it in Yahoo uploader,

this is the link of it : http://d.top4top.net/p_86ub1h1.png

m = 4.00 kg

theta = 30

v = 1 m/s

compression of the spring by the block , x = 0.3 m

d = 5 m

rule : Fk = μ k * N

i must find Fk first to find μ k using conservation energy,

First find N,

N = m x g x cos theta

N = 4 * 9.81 * cos30 = 33.98N

then we find the force to find the spring constant

F = ma

F = 4 * 9.81 = 39.24N

KE = 0.5 mv^2

KE = 0.5 * 39.24 * 1^2 = 19.62J

PE = 0.5 kx^2 => 0.5 Fx

PE = KE

0.5 Fx = 0.5 mv^2

x = v^2 / a = 1^2 / 9.81 = 0.101m

K = f/x = 392.4

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Here i am confused, our professor told us to use mechanical energy and conservation energy to find Fk ,

and the rules is

delta E = delta E mechanics + delta E thermo + delta E internal

K2 + U2 = K1 + U1

The question is long and very tricky, i solve what i can, and i not sure about it actually, if you can help me i will be very thankful.

1 Answer

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  • NCS
    Lv 7
    4 years ago
    Favorite Answer

    I question a couple of your steps:

    (i) F = 4 * 9.81 = 39.24N

    Why do this? This is the weight of the object.

    (ii) KE = 0.5 * 39.24 * 1^2 = 19.62J

    This is wrong. You have put the weight in instead of the mass.

    (iii) delta E = delta E mechanics + delta E thermo + delta E internal

    That looks OK, but

    K2 + U2 = K1 + U1

    has lost something. Instead, you should wind up with

    deltaE = deltaThermo = friction work = initial GPE - final SPE

    since the deltaKE is zero.

    So, start with the horizontal setup to find k:

    spring PE becomes block KE

    ½kx² = ½mv² → ½ cancels

    k * (0.20m)² = 4.00kg * (1.00m/s)²

    k = 100 N/m

    Simple enough.

    Now consider the incline problem:

    initial GPE = friction work + final SPE

    m*g*(5.00m + 0.30m)*sinΘ = µ*m*g*(5.00m + x)*cosΘ + ½k(0.3m)²

    Plug in m, g, k and Θ and solve for µ.

    If you find this helpful, please award Best Answer!

    • NCS
      Lv 7
      4 years agoReport

      You're welcome, and thanks!

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