Post a New Question


posted by .

A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the oscillation is started by elongating the spring 0.15 m and giving the block a speed of 3.0 m/s, then the maximum speed of the block is
A3.7 m/s
B0.18 m/s
C5.2 m/s
D13 m/s
E0.13 m/s

  • Physics -

    PE(max) = KE +PE
    kx(max)²/2 =mv²/2 +kx²/2
    x(max) = sqrt(mv²/k +x²)=
    =sqrt(0.25•9/200 +0.15²)=0.184.
    v(max) = ω•x(max) = x(max) •sqrt(k/m) =
    =0.184•sqrt(200/0.25)=5.2 m/s

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question