Post a New Question

PHYSICS

posted by .

When a 0.116 kg mass is suspended at rest from a certain spring, the spring stretches 3.50 cm. Find the instantaneous acceleration of the mass when it is raised 5.70 cm, compressing the spring 2.20 cm.

  • PHYSICS -

    m•g=k•x₁
    k= m•g/x₁=0.116•9.8/0.035= 32.5 N/m
    When the spring rises 5.7 cm =0.057 m and the spring is compressed to x₂=0.022 m
    F= k•x₂ =ma
    a =k•x₂ /m=32.5•0.022/0.116 =6.16 m/s²

  • PHYSICS -

    .116 * 9.81 = k (.035)
    so
    k = 32.5 N/m

    F = .116*9.18+32.5*.022 = 1.14 + .715 = 1.855 N

    1.855 = .116 a
    a = 16 m/s^2

  • PHYSICS -

    another way
    spring is compressed .057 from equilibrium hanging

    F = k (.057) = 32.5 * .057 = 1.85 N
    a = 1.85/.116 = 16 m/s^2 again

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question