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Physics Question below (springs): Please help!?

A 380g block is firmly attached to a light horizontal spring. The block can slide along the table where the coefficient of friction is 0.30. The block is compressed 18cm and, when released, travels 17cm past the equilibrium point before stopping. What is the spring’s stiffness (spring constant)?

2 Answers

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  • 3 years ago
    Favorite Answer

    Potential energy in spring = work done by friction

    1/2 k x² = F d

    1/2 k (0.18 m)² = (0.380 kg * 9.8 m/s² * 0.30) * (0.18 m + 0.17 m)

    k = 24.1 N/m

  • oubaas
    Lv 7
    3 years ago

    m*g*μ*(x+d) = k/2*x^2

    k = 2*(0.38*9.806*0.30*0.35)/0.18^2 = 24.15 N/m

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