A mass is attached to the end of a spring on a frictionless horizontal surface.

Question

A mass is attached to the end of a spring on a frictionless horizontal surface. We choose a coordinate system where when the spring is at its equilibrium position x 0. The spring is then stretched to a position X0 as shown. The mass is released from rest, and when the spring is at its equilibrium position the speed of the mass is v0 = 2.0 m/s. Next we stretch the spring to a position 2 x0 and release the mass from rest. What is the speed of the mass when the spring is at its equilibrium position?

Explanation / Answer

This is a problem of conservation of elastic energy of the spring restored during our stretching into kinetic energy of the mass when it comes to the equilibrium position
for this first we need to find the spring constant k by solving energy equation:
here , k =spring constant
v= velocity at the equilibrium position
x(1) = streched position in 1st case
x(2) stretched position in the 2nd case
as per given data
x(2) = 2 * x(1) {it is given that it has been stretched twice to the earlier length}
v(1)=2 m/s
v(2)= ??? (being asked)


1/2mv(1)^2 = 1/2kx(1)^2 ---------------1
1/2mv(2)^2 = 1/2kx(2)^2 ---------------2
divinding equation 1 by 2 we get,
v(1)/v(2) = x(1)/x(2)
solving which we get , v(2) = 4 m/s (answer)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.