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
DrRuss/bobpursley/drwls please help!!! - bobpursley, Friday, April 6, 2007 at 4:27pm
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.
DrRuss/bobpursley/drwls please help!!! - Mary, Friday, April 6, 2007 at 8:41pm
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
Thnaks for your help but I figured it out. I was suppose to convert it to m first then change the answer back to cm. Thanks again!
To find the height above the compressed spring from which the block was dropped, we can use the principle of conservation of energy.
1. The potential energy of the block when it is at height h above the compressed spring is given by mgh, where m is the mass of the block (0.30 kg), g is the acceleration due to gravity (9.81 m/s^2), and h is the height we want to find.
2. The potential energy is converted into the potential energy stored in the compressed spring when the block collides with it. The potential energy stored in a spring is given by (1/2)kx^2, where k is the spring constant (450 N/m) and x is the compression of the spring (3.0 cm).
3. Since the block comes to a momentary halt when it collides with the spring, all of its initial kinetic energy is converted into potential energy stored in the spring. Therefore, we can equate the potential energy of the block at height h to the potential energy stored in the compressed spring:
mg(h + x) = (1/2)kx^2
4. Rearranging the equation to solve for h, we have:
h = [(1/2)kx^2 - mgx] / mg
Plugging in the given values:
h = [(1/2)(450 N/m)(0.03 m)^2 - (0.30 kg)(9.81 m/s^2)(0.03 m)] / (0.30 kg)(9.81 m/s^2)
Simplifying the equation, we get:
h ≈ 0.687 m
To convert this to centimeters:
h ≈ 0.687 m * 100 cm/m
h ≈ 68.7 cm
Therefore, the block was dropped from a height of approximately 68.7 cm above the compressed spring.