Posted by **david** on Monday, November 15, 2010 at 12:25am.

A 1.0 kg object is suspended from a vertical spring whose spring constant is 105 N/m. The object is pulled straight down by an additional distance of 0.25 m and released from rest. Find the speed with which the object passes through its original position on the way up.

- Physics help -
**drwls**, Monday, November 15, 2010 at 7:03am
The original position is the equilibrium position, and the velocity is a maximum there. Let that velocity be Vmax

(1/2) M Vmax^2 = (1/2) k X^2

Vmax = X*sqrt(k/M)

k is the spring constant

X = 0.25 m

m = 1.0 kg

I have not considered gravitational potential energy in this derivation. As I recall, its effect cancels out when you define X as being measured from the equilibrium position, with stretching due to the weight taken into account. I am too lazy to verify this.

## Answer This Question

## Related Questions

- Physics - By how much will a 1.52 kg object stretch a spring that is suspended ...
- physics - One end of a massless coil spring is suspended from a rigid ceiling. ...
- physics - One end of a massless coil spring is suspended from a rigid ceiling. ...
- Physics - An object of mass 1.00 kg is attached to a vertical spring with spring...
- physics - An unstretched ideal spring hangs vertically from a fixed support. A 0...
- Physics - A hanging spring stretches by 30.0 cm when an object of mass 490 g is...
- Physics - A hanging spring stretches by 30.0 cm when an object of mass 490 g is...
- PHYSICS - A 0.49 kg object is attached to a spring with a spring constant 167 N/...
- physics help - A 1.0 kg object is suspended from a vertical spring whose spring ...
- Physics - 3 kg object is fastened to a light spring with the intervening cord ...

More Related Questions