Questions on Springs and Simple Harmonic Motion?

Question

1.) A heavy-duty stapling gun uses a 0.173 kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a "ram spring" (k = 37479 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by 3.90*10^-2 m from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by 6.20*10^-3 m when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

2.) 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.

Not sure how to do these; help would be appreciated.

Answer 1

Neither of these needs harmonic motion; they are both dealt with by simple conservation of energy.

The two spring equations are:

Fs = -Cs X
Es = (1/2) Cs X^2

where:
X is the displacement
Cs is the spring constant
Fs is the force exerted by the spring
Es is the energy in the spring

For the second problem, when the object is at rest, the force exerted by the vertical spring must equal the force exerted by gravity. This gives you the initial displacement and the initial energy stored in the spring.

After it is pulled down an additional 0.25 meters, it has additional energy corresponding to the total displacement. When the object returns to its original position, the spring returns to its original displacement, so the extra energy that went into pulling the object down has been converted to kinetic energy:

Ek = (1/2) M V^2

Since you know the energy released by the spring and the mass of the object, you can compute the velocity.

The first problem is even simpler because you are given both displacements explicitly, but the idea is the same.