Understanding Spring Constant: 8.0kg Stone Compression

[JFlash]

The first part of this problem says that a 8.0kg stone rests on top of a spring, and compresses it by 10.0cm. It then asks what the spring constant is. This is the work I did (assuming g = 10m/s^2, not 9.8m/s^2):

(8.0kg)(10m/s^2) = 1/2k(10cm)^2; 80N = 50cm; k = 1.6

But, the textbook says the answer is .8. I don't understand why it's half my answer. Any help would be greatly appreciated.

 

[CompuChip]

On the left hand side, you have the gravitational force,
$$F_g = m g$$
On the left hand side, you have the spring potential energy,
$$W_s = \frac12 k x^2$$
Try taking the force exerted by the spring as well
$$F_s = k x$$

Actually, you could have noticed this mistake by looking at the units... you wrote down 80N = 50cm. Even if you meant
80 N = 50 cm^2 * k (N/cm)
you would have equated N to N cm.
Try to train yourself in spotting these mistakes: unless you know how to convert Newtons into Newtons times centimeters, an alarm bell should go off automatically.

 

[ogb p]

Hello,

Thank you for sharing your work and question. It seems that there may be a misunderstanding of the units used in the problem. The spring constant, k, is typically measured in units of Newtons per meter (N/m), while in your calculation, you used the compression distance in centimeters (cm). In order to get the correct answer, you need to convert the compression distance to meters:

10cm = 0.1m

Now, let's redo the calculation:

(8.0kg)(10m/s^2) = 1/2k(0.1m)^2; 80N = 0.005k; k = 16,000N/m

As you can see, the answer is significantly different from your initial calculation. This is because the compression distance was not converted to the correct unit.

Now, if we divide the answer by 100 (since 1m = 100cm), we get:

16,000N/m ÷ 100 = 160N/cm

This is equivalent to 1.6N/cm, which is half of the textbook's answer of 0.8N/cm. Therefore, the textbook's answer is correct.

I hope this helps clarify the discrepancy between your answer and the textbook's answer. It is important to pay attention to units when solving problems in science, as they can greatly affect the final answer.

Best of luck in your studies!

Sincerely,