[Phys-L] Re: Spring question

[clement at HAL-PC.ORG (John Clement)]

This particular problem brings up a common misconception by students.  If
you ask them what is the maximum stretch of a spring after you hang a block
on it, they will usually give you solution 1, but solution 2 is the
appropriate answer.  They think that the block stops moving when the forces
balance, but they fail to consider that the net force is always downward
until the forces balance, and then it becomes negative as the spring
stretches more.  The upward force then is needed to slow down the block and
stop it at the maximum stretch.

I have seen this confusion in physics teachers during a workshop, so it is
not a trivial misconception.  But it is a misconception that needs to be
addressed.

Solution 1 is used to calculate the equilibrium position after the block has
stopped oscillating, but solution 2 gives the maximum stretch, assuming the
frictional forces are small enough that they can be ignored for one cycle of
oscillation.

These two solutions are actually a fairly good way to see if students
understand forces and energy.

John M. Clement
Houston, TX

> A vertical spring (spring constant k) is hanging freely.
> A block of mass m is attached to the spring and the spring
> streches. At the equilibrium position of the block, where it hangs
> motionless, the spring has streched by x. Determine the mass of the block.
>
> Solution 1:
>
> At the equilibrium the net force on the block is zero
> and hence kx = mg which gives m = kx/g.
>
> Solution 2:
>
> The change in gravitational potential energy of the block
> is mgx. This energy must be in the mass-spring system and
> equals 1/2kx*x (assuming no friction in the spring).
> But this calculation gives m = 1/2* kx/g!!