Question

A vertical spring with k = 490 N/m is standing on the ground. You are holding a 5.8 kg block just above the spring, not quite touching it.

(a) How far does the spring compress if you let go of the block suddenly?
m

(b) How far does the spring compress if you slowly lower the block to the point where you can remove your hand without disturbing it?
0.116 m is the answer I got and know how to get

(c) Why are your two answers different?

I think it is because one of them includes kinetic energy and the other doesn't

PLEASE USE MY VALUES AND EXPLAIN ANSWER! Do not answer if you are not sure...

Answer 1

In the first case we must examine the potential energy being converted into the potential energy of the spring and in the second case, since the process occurs so slowly we can examine the forces acting the mass after it finds equilibrium.

In the first case, we have:

Ui + Si = Uf + Sf // Energy conservation theorem between gravitational potential and the spring potential.

Now lets imagine the x-axis increasing upward with the potential energy of the mass at the start (Ui) we will take as zero since it starts at the origin.

Now, Si is also equal to zero because it is at rest length.

Uf = - m*g*x and the Sf = (1/2) k * x^2

we add the two terms together and get:

0 = x* (0.5 * k*x - m*g) , now cancel the x outside because that reveals a ridiculous answer.

We now have:

x = 2 * m * g / k // after we move around the variables.

In the second case:

sum of the forces = weight of the mass + force of spring keeping it up.

So ...

F = 0 = - m*g*x + k*x

after shifting variables:

x = m * g / k


SO the distance compressed in the free fall of the mass is twice as great.

and

The last portion is the easiest to answer, so we'll start there. The two values will be different because of the kinetic energy of the block. In Case "A" when the block is dropped, it will compress the spring past the equilibrium point and then bounce back again ... oscillating back and forth until it stops (or doesn't in a frictionless environment). In Case "B" the block is lowered to the equilibrium point and left there.

The simplest way to solve the entire question is to figure out the equilibrium point (or solve for case "B").

The 5.8kg block exerts a force of 49N (5.8kg*9.8m/s^2).
The spring has a k-value of 490N/m, so at 0.1m of compression it will exert 49N of force (the equivalent of the block).

As for the maximum compression of the spring in Case "A", my gut says that it's 0.2m, but I can't for the life of me remember quite how to calculate it at this point. It has something to do with calculating the energy ... at the equivalence point, the potential energy of the block (within the system) has been fully converted into kinetic energy. Then for the block to stop, the same kinetic energy must be converted back into potential energy (just in the other direction).