Question

A 50.0 g object is attached to a horizontal spring with a force constant of 5.0 N/m and released from rest with an amplitude of 20.0cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless?

(please write out formula used)

Answer 1

We need to convert . 50.0 g = 0.0500 kg; 20.0 cm = 0.200 m

Consider how much spring potential energy you need to add when you pull the object from its equilibrium position to its 'start position' just before you let it go.

spring potential energy (E) is given by:
E = ½kx²

where k is the spring constant and x is the displacement.
Plugging in values:
E = ½ * 5.0 * 0.200²
E = 0.100 J
the spring potential energy when the object is halfway back to the equilibrium position is given by  

E' = ½ * 5.0 * 0.05²

E' = 0.00625J

So the spring has lost ( 0.100 - 0.00625) = 0.09375 J of spring potential energy.
Energy is conserved. We know it hasn't been lost to friction, so it has all been changed into the kinetic energy of the object.

K.E = ½mv²

where m is the mass and v the velocity.
we now know the k.e. = 0.09375 J, and the mass, so we can find v

0.09375 = ½ * 0.0500 * v²
v² = 3.75
v = 1.936 m/s