Question

A uniform bar is supported by a long uniform negligible mass spring as shown in figure 1. The spring in figure 1 is stretched 3.00 cm from its equilibrium position. The bar and the spring are each cut into three equal pieces and assembled as shown in figure 2. What is the total stretching distance in all the pieces of the springs in figure 2?

Answer 1

Figures are shown below.

Answer 2

Probably 2.00 cm.
Work out stretching of each spring.

Answer 3

We tried a variety of ways. Our first attempt was using 3K (Spring constant) but our teacher said it was incorrect. This method gave us 2.00 cm though, so I'm not sure what we did wrong.

Answer 4

This was my process:
Kt = original k of the uncut spring, supporting the total mass
k = k of each spring, which should be equal as they are the same length

Kt = 3k
Fnet = Sum of All Forces
0 = Fg-Fs m = 3k(x)/9.8
mg = Fs
mg = k(x) Combine both formulas.
m = k (0.03)/9.8

1) TOP MASS/SPRING

k(0.03)/9.8 = 3k(x)/9.8
0.03 = 3x
0.01 = x

2) MIDDLE MASS/SPRING

k(0.03)/9.8 = 9kx/2*9.8
0.03 = 9x/2
x = 1/150

3) BOTTOM MASS/SPRING:
k(0.03)/9.8 = 9kx/9.8
x = 1/300

Therefore, when you add up the three stretches, it gives you 0.02m or 2 cm.

Answer 5

Not sure if that is correct though.

To find the separate equations for each spring/mass, we modified: m = 3k(x)/9.8 such that for the middle spring, m was substituted with (2/3)m (since it only holds 2/3 of the original mass). And the m in for the bottom spring was substituted with (1/3)m (since it holds 1/3 of the original mass).

This gave us the 'm=9kx/2*9.8' as seen in the calculations for spring 2, and 'm=9kx/9.8' as seen in the calc. for spring 3.

Answer 6

Of course, this is all correct. All springs have a 3k spring constant.
The top one holds mass m
The middle one holds mass 2m/3
The bottom one holds mass m/3

Overall stretching is 2.0 cm

Answer 7

Okay, thank you! Just wanted to confirm that.

Answer 8

SO WE WERE RIGHTISH! Yayayayayay.