Ask questions and get helpful answers.

An ideal massless spring is fixed to the wall at one end. A block of mass M attached to the other end of the spring oscillates with amplitude A on a frictionless, horizontal surface. The maximum speed of the block is V_m. The force constant of the spring is

a)Mg/A
b)MgV_m/2A
c)MV_m^2/2A
d)MV_m^2/A^2
e)MV_m^2/2A^2

I have narrowed down the choses to either c) d) or e). I don't really get this question.

The max KE of the block is 1/2Mv^2. The max PE of the spring is 1/2 k A^2
set them equal, and solve for k.

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩

1 answer

  1. Because of conservation of energy, kinetic energy K is equal to the potential energy of the spring: K = U. So (mv^2)/2 = (kA^2)/2. Solving for k, we have k = (mv^2)/A^2, which is d).

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Answer this Question

Related Questions

Still need help?

You can ask a new question or browse existing questions.