Can you please clarify your response to my post Wednesday April 4th at 8:53pm and 8:54pm.
Thanks for all of your help!!!
pleasde repost it.
Yes, repost it. Go to it, copy, and paste it here. Thanks.
A vertical spring with a spring constant of 450 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.30 kg block is dropped from rest. It collides with and sticks to the spring, which is compressed by 3.0 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in cm) above the compressed spring was the block dropped?
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Physics - bobpursley, Wednesday, April 4, 2007 at 9:07pm
The spring compression had 1/2 k x^2 of work done on it. Assuming no losses to friction, then the energy that went into it was mg(h+x). Calculate h
So what is the question?
I lead you to mg(h+x)= 1/2 kx^2
solve for h. If questions, you have to ask specific questions.
Please tell me where I went wrong.
mg (h+x) = 1/2 K x^2
h = 1/2 (450)(3.0)^2/0.30kg x 9.81
h = 687.054
Please disregard. I figured it out. Thanks!
It seems that you have already figured out the answer to your question. However, I can help explain the correct approach for solving the problem.
To determine the height from which the block was dropped, we can apply the principle of conservation of mechanical energy. Assuming no energy losses to friction or air resistance, the total initial energy of the block is equal to the total final energy when it comes to a halt.
The initial energy is given by the potential energy when the block is at the initial height h, which is mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height.
The final energy is the potential energy stored in the compressed spring, which is given by the formula 1/2kx^2, where k is the spring constant and x is the compression of the spring.
So, we have the equation: mgh = 1/2kx^2
In your calculation, you correctly substituted the known values for m, g, k, and x into the equation. However, you made a mistake in the calculation. Let's correct it:
h = (1/2)(450 N/m)(0.03 m)^2 / (0.30 kg)(9.81 m/s^2)
= (1/2)(450)(0.0009) / (0.30)(9.81)
≈ 0.046 m
Since the question asks for the height in centimeters, we need to convert 0.046 m to centimeters by multiplying it by 100:
h ≈ 4.6 cm
Therefore, the block was dropped from a height of approximately 4.6 cm above the compressed spring.
I hope this clears up any confusion you may have had. Let me know if you need further assistance!