Post a New Question


posted by .

A mass of 6.10 kg rests on a smooth surface inclined 39.0o above the horizontal. It is kept from sliding down the plane by a spring attached to the wall. the spring is aligned with the plane and has a spring constant of 126 N/m. How much does the spring stretch?

  • Physics -

    can you show the steps to how to answer the question as well ? just the steps would be fine :)

  • Physics -

    vector m•a=vector m•g +vector F(spring)
    a=0 =>
    0=vector m•g +vector F(spring)
    x: 0 = m•g•sinα –k•x,
    x=m•g•sinα/k=6.1•9.8•sin 39°/126 =0.3 m

  • Physics -

    Fm = M*g = 6.10 * 9.8 = 59.8 N. = Force of the mass.

    Fp = 59.8*sin39 = 37.63 N. = Force parallel to the incline = Force applied to the spring.

    d=37.63/126 * 1m = 0.30 m. = Distance the spring is stretched.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question